Evaluation of special exponential integrals

$$I(n)=\int_{-\infty}^\infty x^{n} e^{-x} e^{-e^{-x}-x}dx$$ Let $u=e^{-x}$, $x=-\ln(u)$, $du=-e^{-x}dx$ $$I(n)=-\int_\infty^0 (-\ln(u))^{n} ue^{-u}du=(-1)^n \int_0^\infty \ln^n(u)u e^{-u}du$$

Now the Gamma function is defined as $$\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$$ So $$\left(\frac{d}{ds}\right)^n\Gamma(s)=\int_0^\infty \ln^n(x) x^{s-1} e^{-x}dx$$ Thus $$I(n)=(-1)^n \left(\frac{d}{ds}\right)^n\Gamma(s)|_{s=2}$$ Now let $n=1,2$ to get the values of the two original integrals. It could help to know that $$\Gamma'(2)=-\gamma+H_1=1-\gamma$$ and that $$\psi_1(2)=\frac{\pi^2}{6}-1$$ where $\psi_1(s)$ is the Trigamma Function.

EDIT: $$\psi_1(s)=\left(\frac{d}{ds}\right)^2 \ln(\Gamma(s))=\frac{\Gamma(s)\Gamma''(s)-\left(\Gamma'(s)\right)^2}{\Gamma^2(s)}$$ Evaluating at $s=2$ we get $$\Gamma''(2)-\left(\Gamma'(2)\right)^2=\frac{\pi^2}{6}-1$$ So $$\Gamma''(2)=\frac{\pi^2}{6}-1 +\left(\Gamma'(2)\right)^2=\frac{\pi^2}{6}-1 +(1-\gamma)^2=\frac{\pi^2}{6}-2\gamma+\gamma^2$$