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}
For more information check answers of my question here
For more clarification according to the answer of @Nikunj, try $n=2$ using this steps in general: Let $$I(a) = \int_0^\infty e^{-nx}x^a\,dx$$ $$\implies \frac{d^nI(a)}{da^n} = \int_0^\infty e^{-nx}x^a(\ln x)^n\,dx$$ Put $nx \rightarrow v$ in the first integral to get: $$I(a) = \frac1{n^{1+a}}\int_0^\infty e^{-v}v^a\,dv$$ $$\implies I(a) = \frac{\Gamma(1+a)}{n^{1+a}}$$ Now $$\implies \frac{d^nI(a)}{da^n}\bigg|_{a=0} = \frac{d^n}{da^n}\left(\frac{\Gamma(1+a)}{n^{1+a}}\right)\bigg|_{a=0}$$ Which evaluates to: $$\frac1{n}\sum_{k=0}^n(-1)^k\binom{n}{k}\Gamma^{(n-k)}(1+a)\ln^k(n)\bigg|_{a=0}$$ Where $\Gamma^{(n-k)}(1+a)$ is the $(n-k)$th derivative of the Gamma function.