= Solution
For a <finitely generated module> $V=\bigoplus_dV_d$ over the <graded ring> $S=k[X_1,\ldots,X_n]$, with each $X_i$ of degree one, its <Poincare series of a graded module> is
$$
P_V(t)=\sum_d\dim_k(V_d)t^d.
$$
This is also its <Hilbert series>. One usually takes $d\ge0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\ldots,a_h$ give a surjection $\bigoplus_j S(-a_j)\to V$, where the <graded shift> is defined by $V(-a)_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional <vector space> and $V_d=0$ for all sufficiently negative $d$.
The <Hilbert-Serre theorem> in the standard grading says
$$
\boxed{P_V(t)=\frac{Q(t)}{(1-t)^n},\qquad Q(t)\in\mathbb Z[t,t^{-1}].}
$$
For a nonnegatively graded $V$, the numerator is in $\mathbb Z[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.
We prove the <Hilbert-Serre theorem> by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional <graded module>; thus $P_V(t)$ is a Laurent <polynomial>. Suppose $n>0$ and put
$$
K=0:_VX_n=\{v\in V:X_nv=0\},\qquad C=V/X_nV.
$$
The <Hilbert basis theorem> makes $S$ a <Noetherian ring>, so $K$ is a <finitely generated module>, as is $C$. Both are <graded modules> killed by $X_n$, hence finitely generated over $S'=k[X_1,\ldots,X_{n-1}]$.
The graded <exact sequence>
$$
0\longrightarrow K(-1)\longrightarrow V(-1)
\xrightarrow{\ X_n\ }V\longrightarrow C\longrightarrow0
$$
identifies the <kernel> and <cokernel> of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yields
$$
tP_K(t)-tP_V(t)+P_V(t)-P_C(t)=0,
$$
so
$$
(1-t)P_V(t)=P_C(t)-tP_K(t).
$$
By induction the right side has denominator $(1-t)^{n-1}$, which proves the required denominator $(1-t)^n$. Since $(1-t)^nP_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent <polynomial> numerator is an ordinary <polynomial> in that case.
For $n\ge1$, this also implies the usual <Hilbert polynomial> consequence. Writing $Q(t)=\sum_jq_jt^j$ and using
$$
\frac1{(1-t)^n}=\sum_{d\ge0}\binom{d+n-1}{n-1}t^d
$$
gives, for all sufficiently large $d$,
$$
\dim_kV_d=\sum_jq_j\binom{d-j+n-1}{n-1},
$$
a <polynomial> in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.
For the final request, a <free resolution> of $V$ is an <exact sequence>
$$
\cdots\longrightarrow F_2\xrightarrow{d_2}F_1\xrightarrow{d_1}F_0\longrightarrow V\longrightarrow0
$$
with each $F_i$ a <free module>. In the graded setting we take finite <direct sums> of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The <syzygy modules> are the successive <kernels> which record the relations among generators, then relations among those relations, and so on.
Put $\mathfrak m=(X_1,\ldots,X_n)$, the homogeneous <maximal ideal> with $S/\mathfrak m=k$. Choose a homogeneous $k$-basis of $V/\mathfrak mV$ and lift it to $V$. These lifts generate $V$ by the <graded Nakayama lemma>: a bounded-below <graded module> $W$ satisfying $W=\mathfrak mW$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the <submodule> generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an <isomorphism> modulo $\mathfrak m$.
Its <kernel> is finitely generated because $S$ is a <Noetherian ring>. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a <minimal graded free resolution>, meaning
$$
d_i(F_i)\subseteq\mathfrak mF_{i-1}\qquad(i\ge1).
$$
Indeed, at each stage the free cover induces an <isomorphism> modulo $\mathfrak m$, so its kernel lies in $\mathfrak m$ times its source. The sequence is exact by construction, though it may at first appear infinite.
The <Koszul complex> on $X_1,\ldots,X_n$ is
$$
K_i=\bigwedge^i S^n\cong S(-i)^{\binom ni},\qquad 0\le i\le n,
$$
with basis vectors $e_j$ given degree one, and differential
$$
\partial(e_{j_1}\wedge\cdots\wedge e_{j_i})
=\sum_{a=1}^{i}(-1)^{a-1}X_{j_a}
e_{j_1}\wedge\cdots\wedge\widehat{e_{j_a}}\wedge\cdots\wedge e_{j_i}.
$$
The hat means omission of that factor. Every pair of terms in $\partial^2$ cancels with opposite signs, so this is a <chain complex>. The sequence $X_1,\ldots,X_n$ is a <regular sequence>: after quotienting by the first $j-1$ variables, the next variable is a <non-zero-divisor> in the remaining <polynomial ring>.
The <Koszul complex> is consequently exact in positive degrees and has $H_0=k$, so it is a <Koszul resolution> of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\xrightarrow{X_1}S\to S/(X_1)\to0$. Appending $X_j$ constructs the <mapping cone> of multiplication by $X_j$ on the previous <chain complex>. Its <homology> is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\ldots,X_{j-1})$; its degree-zero <homology> is the further quotient. Here a <mapping cone> combines a <chain complex> with a shifted copy and adds the given multiplication map to the differential, thereby measuring its <kernel> and <cokernel> on <homology>.
The <Tor functor> $\operatorname{Tor}_i^S(V,k)$ is defined as the degree-$i$ <homology> of $F_\bullet\otimes_S k$ for a <free resolution> $F_\bullet$ of $V$. It can equally be computed as the <homology> of $V\otimes_S K_\bullet$, using the <Koszul resolution> of $k$. To see the equality, form the <double complex> $F_\bullet\otimes_S K_\bullet$. Taking <homology> first along $F_\bullet$ leaves $V\otimes_S K_\bullet$, since each $K_i$ is a <free module>; taking <homology> first along $K_\bullet$ leaves $F_\bullet\otimes_S k$, since each $F_i$ is free. The finite sums along each total degree identify both with the <homology> of the total complex. This also explains the symmetry used in computing the <Tor functor>.
Since $K_i=0$ for $i>n$, we have $\operatorname{Tor}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the <minimal graded free resolution> become zero after tensoring with $k=S/\mathfrak m$, so
$$
\operatorname{Tor}_i^S(V,k)\cong F_i\otimes_S k=F_i/\mathfrak mF_i.
$$
Thus $F_i/\mathfrak mF_i=0$ for $i>n$, and the <graded Nakayama lemma> implies $F_i=0$ for those $i$. We have proved the required <Hilbert syzygy theorem>:
$$
\boxed{0\longrightarrow F_n\longrightarrow\cdots\longrightarrow F_0\longrightarrow V\longrightarrow0,}
$$
where some initial $F_i$ may be zero. The argument supplies finitely generated graded <free modules>, which is stronger than merely giving an ungraded <free resolution>.
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