Find $\int_0^1(\ln x)^n\hspace{1mm}dx$ I am not a big fan of induction, it's just a personal preference.
Is there a method other than induction.
Answer is $n!$ by the way 

Let, $\ln x=t\implies \frac{dx}{x}=dt\implies dx=e^tdt$, we have $$\int_{0}^{1}(\ln x)^ndx=\int_{-\infty}^{0}(t)^ne^tdt$$ $$=\int_{0}^{\infty}(-t)^ne^{-t}dt$$ $$=(-1)^n\int_{0}^{\infty}e^{-t}t^ndt$$ Now, using Laplace transform $\color{blue}{\int_{0}^{\infty}e^{-st}f(t)dt=L[f(t)]}$ & $\color{blue}{L[t^n]=\frac{n!}{s^{n+1}}}$, we get  $$(-1)^n\int_{0}^{\infty}e^{-t}t^n dt=(-1)^n L[t^n]_{s=1}$$ $$=(-1)^n\left[\frac{n!}{s^{n+1}}\right]_{s=1}$$ $$=(-1)^n\left[\frac{n!}{(1)^{n+1}}\right]=(-1)^n(n!)$$ $$\implies \color{blue}{\int_{0}^{1}(\ln x)^ndx=(-1)^n(n!)}$$ Let, $n$ be an even integer then we have  $$ \color{blue}{\int_{0}^{1}(\ln x)^ndx=(-1)^n(n!)=n!}$$
