show that $\int_{0}^{1} (-\ln(t))^n dt=\int_{0}^{\infty} e^{-t}t^n dt$
I tried to make a variable change using the change of variables theorem,
with $u=-\ln (t)$ such that $t=e^{-u}$ and $dt=-e^{-u} du$ thus, we obtain something similar to the r.h.s of, but in terms of u. $$\int_{0}^{1} (-\ln(t))^n dt=\int_{e^{-\alpha}=0}^0 (-\ln(e^{-u}))^{n}(-e^{-u} )du\\ =\int_{e^{-\alpha}=0}^0 (\ln(e^u))^n (-e^{-u} )du=\int_{e^{-\alpha}=0}^0 u^n (-e^{-u} )du=-\int_{0}^{e^{-\alpha}=0}u^n (-e^{-u} )du$$ In addition, the upper limit is not indefinite, I suppose that here comes the indefinite integral.
Is this a correct path? If so, what should I try to do afterwards?