Show that $\exp\left(\frac{1}{x}\log\frac{e^{x}-1}{x}\right)$ is increasing.
Write
$$ f(x) = \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} = \left( \int_{0}^{1} e^{xs} \, \mathrm{d}s \right)^{\frac{1}{x}}. $$
Now let $0 < x < y$ be arbitrary and write $p = \frac{y}{x} > 1$. Then by the Jensen's inequality applied to the strictly convex function $\varphi(t) = t^p$ over $[0, \infty)$, we get
$$ f(x)^{y} = \varphi\left( \int_{0}^{1} e^{xs} \, \mathrm{d}s \right) < \int_{0}^{1} \varphi(e^{xs}) \, \mathrm{d}s = f(y)^{y}, $$
and therefore $f(x) < f(y)$ as desired.
Remarks.
This is a particular instance of a more general observation that the $L^p$-norm $$\| X\|_{L^p} := (\mathbb{E}[|X|^p])^{1/p}$$ of a random variable $X$ is non-decreasing in $p$.
We may instead use the Hölder's inequality in the proof.
First \begin{eqnarray} F(x)&:=&e^{x}(x-1)+1-(e^{x}-1)\log(e^{x}-1)+(e^{x}-1)\log x\\ &=&(e^{x}-1)(x-1)+x-(e^{x}-1)\log(\frac{e^{x}-1}{x})\\ &=&(e^x-1)g(x) \end{eqnarray} where $$ g(x)=x-1+\frac{x}{e^x-1}-\log(\frac{e^{x}-1}{x}).$$ Clearly $g(x)=0$. Note $$ g'(x)=\frac{e^{2x}+1-e^x(2+x^2)}{x(e^x-1)^2}=\frac{e^x}{x(e^x-1)^2}(e^x+e^{-x}-2-x^2). $$ It is very easy to show that $$ e^x+e^{-x}-2-x^2>0$$ which I omit the detail and hence $g'(x)>0$ for $x>0$. This gives $g(x)>g(0)=0$. So $F(x)>0$ which implies $f(x)$ is increasing.