= Solution
Use the <determinant-normalized slash operator> and let $\mathcal M_n$ be all integral <matrices> of positive <determinant> $n$. Their left $SL_2(\mathbb Z)$ cosets have the unique <determinant-n matrix representatives for Hecke operators>
$$
\begin{pmatrix}a&b\\0&d\end{pmatrix},\qquad a,d>0,\ ad=n,\ 0\leq b<d.
$$
Integer row operations reduce the first column to $(a,0)^T$, and a row shear then reduces $b$ modulo $d$. Define
$$
\boxed{T_nf=n^{k/2-1}\sum_{\alpha\in SL_2(\mathbb Z)\backslash\mathcal M_n}f|_k\alpha
=n^{k-1}\sum_{ad=n}d^{-k}\sum_{b=0}^{d-1}f\left(\frac{az+b}{d}\right).}
$$
Changing representatives does not affect the sum, while right multiplication by the <modular group> permutes the cosets. Hence $T_nf$ has the same modular transformation law. Rational positive-determinant slash transforms are holomorphic on $\mathbb H$ and preserve <modular cusp> vanishing: choose an integral <matrix> carrying infinity to their rational image <modular cusp>, after which the remaining <matrix> is upper triangular. Its positive dilation sends the original decaying <modular cusp> expansion to another decaying expansion. Thus these are well-defined <Hecke operators> on $S_k$.
For the <Fourier expansion>, insert $f(z)=\sum_{m\geq1}a(m)e^{2\pi imz}$. The sum over $b$ vanishes unless $d\mid m$, in which case it is $d$. Writing $m=d\ell$, the resulting exponential is $q^{a\ell}$, and its factor is $n^{k-1}d^{1-k}=a^{k-1}$. Therefore
$$
\boxed{a(r;T_nf)=\sum_{a\mid(n,r)}a^{k-1}a(nr/a^2;f).}
$$
In particular, the first <coefficient> of $T_nf$ is $a(n;f)$.
Assume $f\ne0$ is a simultaneous <Hecke eigenform>. The first-coefficient identity gives $a(n)=\lambda(n)a(1)$. If $a(1)=0$, all <coefficients> vanish, contradicting $f\ne0$. After dividing by $a(1)$, we may therefore suppose $a(1)=1$, and then $\lambda(n)=a(n)$.
We derive the relations needed for the <Euler product>. For coprime $m,n$, applying the <coefficient> formula twice splits the divisors into their disjoint prime supports and yields $T_mT_n=T_{mn}$. For one prime define formal Fourier operators $U_p(\sum a(r)q^r)=\sum a(pr)q^r$ and $V_pf(z)=f(pz)$. The <coefficient> formula gives
$$
T_{p^r}=\sum_{j=0}^r p^{j(k-1)}V_p^jU_p^{r-j},\qquad U_pV_p=I.
$$
Multiplying this finite sum by $T_p=U_p+p^{k-1}V_p$ shows that
$$
T_pT_{p^r}=T_{p^{r+1}}+p^{k-1}T_{p^{r-1}}\qquad(r\geq1).
$$
These auxiliary operators need not separately preserve level one; the identities concern their action on <Fourier series>. Applied to the eigenform, they give
$$
a(mn)=a(m)a(n)\quad((m,n)=1),\qquad
a(p^{r+1})=a(p)a(p^r)-p^{k-1}a(p^{r-1}).
$$
The prime-power generating function consequently satisfies
$$
\sum_{r\geq0}a(p^r)X^r=\frac1{1-a(p)X+p^{k-1}X^2}.
$$
Multiplicativity and unique <prime factorization> now prove the <Euler product of a Hecke eigenform>:
$$
\boxed{L(f,s)=\prod_p\left(1-a(p)p^{-s}+p^{k-1-2s}\right)^{-1}.}
$$
For the original unnormalized form, the right side is multiplied by $a(1)$ and $a(p)$ in the factors is replaced by $\lambda(p)=a(p)/a(1)$. The zero form has $L=0$ and is excluded from the nonzero-eigenform product assertion.
These manipulations initially hold in an absolutely convergent right half-plane. For example, $y^{k/2}|f(z)|$ is bounded by modular invariance and <modular cusp> decay. The <coefficient> integral at $y=1/n$ gives $|a(n)|\leq Cn^{k/2}$, so $\operatorname{Re}s>k/2+1$ is sufficient. No analytic continuation is needed to justify the initial product.
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