As the question suggests, what is the easiest way to see that$$\zeta\_{\mathbb{Z}[i]}(s) = \zeta(s)L(s, \chi)?$$Here, $\chi$ is the homomorphism $(\mathbb{Z}/4\mathbb{Z})^\times \to \mathbb{C}^\times$ which sends $3$ mod $4$ to $-1$ and $L(s, \chi)$ is the Dirichlet $L$-function of $\chi$.



Easiest way to see that $\zeta_{\mathbb{Z}[i]}(s) = \zeta(s) L(s, \chi)$?
As mentioned in the comment, the trick is to look in the Euler product (which holds in the half plane of $Re(s)$ large enough and then extend by analytic continuation). Since it's an arithmetic identification, one should look at the behavior of primes in $\mathbb{Z}$ splitting in $\mathbb{Z}[i]$. 


It's not a hard fact to show that $2$ is the only ramified prime in $\mathbb{Z} \subset \mathbb{Z}[i]$ (the discriminant of this number field is $4$). It also follows via the following formula $\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}{2}}$ that an odd prime splits if and only if it reduces to $1 \pmod 4$.


Let's unfold the RHS, the product of (dimension $1$) $L$-functions, first looking at odd primes $p \equiv 1 \pmod 4$, so that $p = \mathfrak{p} \mathfrak{p}'$ in $\mathbb{Z}[i]$. The local factor at $p$ on the RHS is given by $\left(1 - p^{-s} \right)^{-2}$. The corresponding local factor on the LHS is given by $\left(1 - N(\mathfrak{p})^{-s}\right)^{-1} \left(1 - N(\mathfrak{p}')\right)^{-1}$, where $N$ is the norm map given by the extension $\mathbb{Q}(i)/\mathbb{Q}$. But $N(\mathfrak{p}) = N(\mathfrak{p}') = p$.


We unfold a similar argument for other odd primes $p$ that remain inert in $\mathbb{Z}[i]$. The local factor at $p$ on the RHS is given by $\left(1 - p^{-s}\right)^{-1} \left(1 + p^{-s} \right)^{-1} = \left(1 - p^{-2s}\right)^{-1}$. On the RHS, the local factor corresponding to $p$ is given by $\left(1 - N(p)^{-s}\right)^{-1}$. But $N(p) = p^2$ if $p$ is inert.


Lastly, for $p = 2$, we have $2 = \mathfrak{q}^2$ in $\mathbb{Z}[i]$. Because the character vanishes at the place $2$, the local factor on the RHS lives exclusively in $\zeta(s)$, and is thus $\left(1 - 2^{-s}\right)^{-1}$. But the LHS has a corresponding local factor of $\left(1 - N(\mathfrak{q})\right)^{-1}$, and $N(\mathfrak{q}) = 2$.


Just for completeness, convergence of our infinite Euler products is not an issue. 


Just as a brief remark (up to you how seriously to take this for now), although these functions are analytic manifestations of arithmetic gadgets, these factorization problems are, in nature, more algebraic/arithmetic than analytic. In particular, $L$-functions given by continuous, char. $0$ Galois representations illustrate this theme well (decomposing representations is given by factoring corresponding $L$-functions, which is crucial to showing such an $L$-function is given by a product of Hecke $L$-functions).


