Hint: you may try $n=2$ and follow in general this :\begin{align} \int_0^\infty \left(\frac{\log(x)}{e^x}\right)^n\,dx&=\int_0^\infty e^{-nx}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^n(x)\,dx\\\\ &=\frac1n\sum_{k=0}^n\binom{n}{k}\log^{n-k}(n) \left.\left(\frac{d^n \Gamma(x+1)}{dx^n}\right)\right|_{x=0} \end{align}
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