= Solution
Put $Z_n=n(1-M_n)$. Its <survival function> is
$$
\mathbb P(Z_n>z)=\begin{cases}
1,&z<0,\\
(1-z/n)^n,&0\leq z\leq n,\\
0,&z>n.
\end{cases}
$$
For each fixed $z\geq0$, eventually $n\geq z$, and the <exponential limit> gives $(1-z/n)^n\to e^{-z}$. Thus its <cumulative distribution function> tends to $0$ for $z<0$ and to $1-e^{-z}$ for $z\geq0$. This continuous limiting <cumulative distribution function> identifies the <exponential distribution> with rate one:
$$
\boxed{n(1-M_n)\xrightarrow{\mathrm d}\operatorname{Exp}(1).}
$$
This is the <exponential endpoint limit of a uniform maximum>: the gap from the upper endpoint is typically of order $1/n$, although the maximum itself approaches a constant.
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