= Solution
For <compact> <connected> oriented $n$-manifolds without boundary, the degree is characterized by
$$
\int_MF^*\eta=\deg(F)\int_N\eta
$$
for every smooth top-degree form $\eta$. Equivalently, at any <regular value> $y$,
$$
\deg(F)=\sum_{x\in F^{-1}(y)}\operatorname{sign}\det D F_x.
$$
The <preimage> is finite by <compactness> and the <inverse function theorem>. Taking a form supported in a small neighbourhood of a <regular value> and applying change of variables proves the signed-count formula. The top <de Rham cohomology> of a <compact> <connected> oriented <manifold> is one-dimensional, with integration giving its coordinate; <pullback> and <Stokes theorem> make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the <regular value>. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under <homotopy>, as follows by applying <Stokes theorem> to the <pullback> on $M\times[0,1]$.
For the <polynomial> assertion take $k\ge1$. The map $P:\mathbb C\to\mathbb C$ extends smoothly to a map of the <Riemann sphere> fixing infinity: in reciprocal coordinates its target coordinate is
$$
\frac1{P(1/\zeta)}=\frac{\zeta^k}{1+a_1\zeta+\cdots+a_k\zeta^k}.
$$
The <homotopy> $P_t(z)=z^k+t(a_1z^{k-1}+\cdots+a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the <regular value> one, each <orientation>-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\deg P=k\ne0$. It is onto the sphere and must attain zero at a finite point. This proves \b[every nonconstant complex <polynomial> has a root], without assuming the result to count roots of a general <polynomial>.
For a proper smooth $F:\mathbb R^n\to\mathbb R^n$, choose a compactly supported $n$-form $\eta$ with $\int\eta=1$ and define
$$
\deg F=\int_{\mathbb R^n}F^*\eta.
$$
Properness makes its <pullback> compactly supported. If $\eta'$ is another choice, the given result gives $\eta-\eta'=d\beta$ with <compact> support. The difference of the defining integrals is $\int d(F^*\beta)=0$, proving <independence>. Sard's theorem supplies a <regular value>. A <proper map> has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-<function> neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an <integer>.
Finally $x\mapsto x^k$ is proper for $k\ge1$. For odd $k$, a positive <regular value> has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. Consequently
$$
\boxed{\deg(x\mapsto x^k)=\begin{cases}1,&k\ge1\text{ odd},\\0,&k\ge2\text{ even}.\end{cases}}
$$
For $k=0$ it is the constant map one, whose inverse image of the <compact> singleton $\{1\}$ is all of $\mathbb R$. It is \b[not proper], so its degree is not defined by the proper-map construction requested here.
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