It is well known that $$ \int_{0}^{1}\ln^p(1/x)\,dx=\Gamma(p+1) $$ In particular, if $p=n$ is a positive integer, then $$ I_n:=\int_{0}^{1}\ln^n(1/x)\,dx=(n+1)! $$ Therefore, we obtain the following recursive relation $$ I_{n+1}=(n+1)I_n\quad;\quad I_1=1 $$ Is there an easy way, using only elementary methods, to establish that relation?
My attempt: Using integration by parts \begin{align} I_{n+1}&=\left[\ln^n\left(\frac{1}{x}\right)x\left(1+\ln\left(\frac{1}{x}\right)\right)\right]_{0}^{1}\\ &\qquad-\int_{0}^{1}(n+1)\ln^n\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)x\left(1+\ln\left(\frac{1}{x}\right)\right)\,dx \end{align} How do I continue from here ?
$$I_{n+1}=\int_0^1 1 \log ^{n+1}\left(\frac{1}{x}\right) \, dx=\left[x \log ^{n+1}\left(\frac{1}{x}\right)\right]_0^1- \int_0^1 -\frac{x \left((n+1) \log ^n\left(\frac{1}{x}\right)\right)}{x} \, dx=$$ $$=\left[0-\underset{x\to 0}{\text{lim}}\;x \log ^{n+1}\left(\frac{1}{x}\right)\right]+(n+1)I_n=0+(n+1)I_n$$ $$I_{n+1}=(n+1)I_n$$