Let $X\sim \mathcal{U}(0,1)$ and $Y = \ln\left((1-X)^{-c}\right)$ for some $c>0$.
We need to show that $Y \sim {\mathrm{Exp}}(1/c)$.
We got to here but couldn't go further:
$ \mathbb{P}(Y \leqslant y) = \mathbb{P}( X \leqslant 1- \mathrm{e}^{-c y}) $
Thanks in advance.