Show that the integral of $1/(1-x^x)$ from $0$ to $1/e$ is divergent How can you show that $\int_0^{1/e}\frac{dx}{1-x^x}$ diverges? Do you have to substitute $x = \frac1u$?

Based on the answer by DeepSea I have written down the complete proof.
This proof makes use of the following facts:
$$e^{x} \ge 1+x \tag{1}\label{1}$$ $$\int_1^{+\infty}\frac{dx}{x} = +\infty \tag{2}\label{2}$$
Consider \eqref{1} for $-\infty < x < 0$ in particular:
$$\begin{align} e^{x} & \stackrel{\eqref{1}}> 1+x \\ -x & \gt 1-e^x \\ \frac{1}{1-e^x} & \gt -\frac1x \tag{3}\label{3}\end{align}$$
Use \eqref{3} and the integral monotonicity to show:
$$\begin{align} \int_0^{1/e}\frac{dx}{1-x^x} & = \int_0^{1/e}\frac{dx}{1-e^{x\ln{x}}} \\ & \stackrel{\eqref{3}}> -\int_0^{1/e}\frac{dx}{x\ln{x}} \tag{4}\label{4}\end{align}$$
Substitute $x$ with $e^{-u}$ $(dx = -e^{-u}du)$:
$$\begin{align} \int_0^{1/e}\frac{dx}{x\ln{x}} & = \int_{+\infty}^{1}\frac{-e^{-u}}{e^{-u}(-u)}du \\ & = -\int_1^{+\infty}\frac{du}{u} \tag{5}\label{5}\end{align}$$
Now, combine \eqref{2}, \eqref{4} and \eqref{5} to proof that $\int_0^{1/e}\frac{dx}{1-x^x} = +\infty$ (and thus diverges):
$$\begin{align} \int_0^{1/e}\frac{dx}{1-x^x} & \stackrel{\eqref{4}}> -\int_0^{1/e}\frac{dx}{x\ln{x}} \\ & \stackrel{\eqref{5}}= \int_1^{+\infty}\frac{du}{u} \\ & \stackrel{\eqref{2}}= +\infty \end{align}$$
This concludes the proof.
