= Solution
Because $Z=e^{-Y}>0$, its <cumulative distribution function> is zero for $z\le0$. For $z>0$, the transformation is decreasing, and $Y$ has a continuous <cumulative distribution function>, so
$$
\mathbb P(Z\le z)=\mathbb P(Y\ge-\ln z)=1-F(-\ln z)=1-e^{-z}.
$$
Thus the <exponential transform of a standard Gumbel variable> gives
$$
\boxed{F_Z(z)=\begin{cases}0,&z\le0,\\1-e^{-z},&z>0,\end{cases}\qquad Z\sim\operatorname{Exp}(1).}
$$
This is the <exponential distribution> with rate $1$.
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