$~\quad~$ Whatever the real reason might have been (or if indeed it was even a matter of conscious choice to begin with) I'm afraid we may never truly know. However, I do find it a fortunate coincidence, inasmuch as it enables many mathematical results to be expressed in terms of the $\Gamma$ and $\zeta$ or $\eta$ functions of the same argument. Here are but a few beautiful examples:
$$\int_0^\infty\frac{x^z}{e^x-1}dx=\Gamma(z+1)~\zeta(z+1),\qquad\qquad\qquad\qquad\Re(z)>1.$$
$$\int_0^\infty\frac{x^z}{e^x+1}dx=\Gamma(z+1)~\eta(z+1),\qquad\qquad\qquad\qquad\Re(z)>1.$$
$$\oint_\gamma\frac{(-x)^z}{e^x-1}dx=2i~\Gamma(z+1)~\zeta(z+1)\sin(z\pi),\qquad\qquad\qquad z\in\mathbb C,$$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ where $\gamma=R~e^{i\alpha}$, with $\alpha\in(0,2\pi)$ and $R\to\infty$.
$$\int_0^\infty\ln\Big(1-e^{-x^{\Large a}}\Big)dx=-\Gamma\bigg(\frac1a+1\bigg)~\zeta\bigg(\frac1a+1\bigg),\qquad\qquad a>0.$$
$$\int_0^\infty\ln\Big(1+e^{-x^{\Large a}}\Big)dx=\quad\Gamma\bigg(\frac1a+1\bigg)~\eta\bigg(\frac1a+1\bigg),\qquad\qquad a>0.$$
$$\ln\Gamma(x)=\zeta'(0,x)-\zeta'(0),\qquad\qquad\qquad\qquad\qquad\qquad x\in\mathbb R^\star\setminus\mathbb Z^-.$$