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 $\chi$.
2 Answers
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 $\left( \frac{1}{p} \right) = (1)^{\frac{p1}{2}}$ 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).
Kevin Yang posted a stellar solution to this problem while I was typing. Here is mine.
We have that$${{\mathbb{Z}[i]}\over{(p)}} \cong \begin{cases} {{\mathbb{Z}[i]}\over{\mathfrak{m}}}\text{ for maximal ideal }\mathfrak{m} \neq \mathfrak{m}_2 & \text{if $p \equiv 1 \text{ }(\text{mod }4)$} \\{{\mathbb{Z}[i]}\over{\mathfrak{m}_1}}{{\mathbb{Z}[i]}\over{\mathfrak{m}_2}} \text{ for maximal ideals }\mathfrak{m}_1 \neq \mathfrak{m}_2 & \text{if $p \equiv 3 \text{ }(\text{mod }4)$}\\ {{\mathbb{Z}[i]}\over{((1+i)^2)}} \text{ for maximal ideal }(1 + i) & \text{if $p = 2$.}\end{cases}$$Therefore,$$\#{{\mathbb{Z}[i]}\over{(p)}} \cong \begin{cases} \#{{\mathbb{Z}[i]}\over{\mathfrak{m}}}\text{ for maximal ideal }\mathfrak{m} \neq \mathfrak{m}_2 & \text{if $p \equiv 1 \text{ }(\text{mod }4)$} \\\#{{\mathbb{Z}[i]}\over{\mathfrak{m}_1}}{{\mathbb{Z}[i]}\over{\mathfrak{m}_2}} \text{ for maximal ideals }\mathfrak{m}_1 \neq \mathfrak{m}_2 & \text{if $p \equiv 3 \text{ }(\text{mod }4)$}\\ \#{{\mathbb{Z}[i]}\over{((1+i)^2)}} \text{ for maximal ideal }(1 + i) & \text{if $p = 2$.}\end{cases}$$So we have that$$\zeta(s)L(s, \chi) = \left(\prod_{p\, \equiv\, 1 \,(\text{mod}\,4)} {1\over{1  p^{s}}}{1\over{1  p^{s}}}\right)\left(\prod_{p\, \equiv\, 1 \,(\text{mod}\,4)} {1\over{1  p^{s}}}{1\over{1 + p^{s}}}\right)\left({1\over{1  2^{s}}}\right).$$Now, we have that$$\zeta_{\mathbb{Z}[i]}(s) = \prod_{\mathfrak{m} \in \text{max}(\mathbb{Z}[i])} {1\over{\left(1  \#{{\mathbb{Z}[i]}\over{\mathfrak{m}}}\right)^{s}}}$$$$= \left({1\over{1  \#\left({{\mathbb{Z}[i]}\over{(1+i)}}\right)^{s}}}\right)\left(\prod_{\mathfrak{m} \in \text{max}(\mathbb{Z}[i])} {1\over{\left(1  \#{{\mathbb{Z}[i]}\over{\mathfrak{m}}}\right)^{s}}}\right)$$$$= \left({1\over{1  2^{s}}}\right)\left(\prod_{p \text{ is prime }\neq\,2} \prod_{\substack{\mathfrak{n} \in \text{max}(\mathbb{Z}[i]),\\ (p) \subseteq \mathfrak{n}}} {1\over{\left(1  \#{{\mathbb{Z}[i]}\over{\mathfrak{n}}}\right)^{s}}}\right)$$$$=\left({1\over{1  2^{s}}}\right)\left(\prod_{p\,\equiv\,1\,(\text{mod}\,4)} \prod_{\mathfrak{n} = (p)} {1\over{1  \#\left({{\mathbb{F}_p[T]}\over{(T^2 + 1)}}\right)^{s}}}\right)\left(\prod_{p\,\equiv\,3\,(\text{mod}\,4)} \prod_{\mathfrak{n} \in \{\mathfrak{m}_1,\mathfrak{m}_2\}} {1\over{1  \#\left({{\mathbb{Z}[i]}\over{\mathfrak{n}}}\right)^{s}}}\right)$$$$=\left({1\over{1  2^{s}}}\right)\left(\prod_{p\, \equiv\, 1\,(\text{mod}\,4)} {1\over{1  p^{2s}}}\right)\left(\prod_{p \,\equiv\,3\,(\text{mod}\,4)} {1\over{(1  p^{s})^{2}}}\right)$$$$=\zeta(s) L(s, \chi),$$as desired.
I want to augment the above with some philosophical reflection. We compare$$\zeta_{\mathbb{Z}[i](s)} = \zeta(s)L(s, \chi)\tag*{(*)}$$and$$\zeta_{\mathbb{Z}[\sqrt[3]{2}]}(s) = \zeta(s)L(s, f).\tag*{(**)}$$$(**)$ is a good example that demonstrates the Langlands correspondence.
We proved $(*)$ above. $\chi$ in $(*)$ is a Dirichlet character $(\mathbb{Z}/4\mathbb{Z})^\times \to \mathbb{C}^\times$ defined by $\chi(1) = 1$, $\chi(3) = 1$. This $(*)$ tells that a prime number $p \neq 2$ decomposes into two if $\mathbb{Z}[i]$ if and only if $p \equiv 1\,(\text{mod}\,4)$.
In $(**)$, we have that$$f(z) = \eta(6z)(18z) = q\prod_{n=1}^\infty (1  q^{6n})(1  q^{18n}),\text{ }q = e^{2\pi iz}$$and$$L(s, f) = \sum_{n=1}^\infty a_nn^{s},$$with $a_n$ determined by$$q =\prod_{n=1}^\infty (1  q^{6n})(1  q^{18n}) = \sum_{n=1}^\infty a_nq^n$$$$=q  q^7  q^{13} q^{19} + q^{25} + 2q^{31}  q^{37} + 2q^{43}  q^{61}  q^{67}  q^{73}  q^{79} + q^{91}  q^{97}  q^{103} \dots$$(The proof of $(**)$ is quite difficult, so we will not include it here.) $(**)$ tells us that for a prime number $p \neq 2, 3$, we have that $(p)$ in $\mathbb{Z}[\sqrt[3]{2}]$ decomposes into a product of three maximal ideals if and only if $a_p = 2$. (It is not very hard to see that $(**)$ implies this, but we will not explain it here.) For example, $31$ decomposes into three in $\mathbb{Z}[\sqrt[3]{2}]$ as$$31 = (3 + 2^{1/3})(3 + 3(2^{1/3}) + 2^{2/3})(3  3(2^{1/3}) + 2^{2/3}).$$This $f$ is a modular form, and this is an example of the Langlands correspondence between arithmetic and modular forms. The Langlands correspondence tells us, roughly speaking, that Hasse zeta functions are expressed by zeta functions of modular forms. (The Dirichlet character is regarded as a special case of a modular form.)

6$\begingroup$ The $\mathbb{Z}[\sqrt[3]{2}]$ example is gorgeous. Do you have a reference for it, and perhaps for other similar modular factorizations of nonGalois $\zeta_K$? $\endgroup$– MyshkinCommented Jul 23, 2015 at 11:25