Let, ${e^{ - x}} = t$
$ \Rightarrow - {e^{ - x}}dx = dt$
$ \Rightarrow - tdx = dt$
$I = \int\limits_1^0 {\frac{{{t^n} - t}}{{ - \ln t}}\frac{{dt}}{{ - t}}} = - \int\limits_0^1 {\frac{{{t^{n - 1}} - 1}}{{\ln t}}dt} $
But, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } - 1}}{{\ln x}}dx = \ln (\lambda + 1)} $ (Proof can be seen in this problem)
So, $I = - \ln (n - 1 + 1) = - \ln n$