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, $I =  – \ln (n – 1 + 1) =  – \ln n$