= Solution
Write $M_n=\max_iE_i$. Each <exponential distribution> has <cumulative distribution function> $1-e^{-x}$ for $x\ge0$. Independence therefore gives the <maximum of independent exponential random variables> formula
$$
\boxed{\mathbb P(M_n\le x)=\begin{cases}0,&x<0,\\(1-e^{-x})^n,&x\ge0.\end{cases}}
$$
For any fixed $y$, the threshold $y+\ln n$ is nonnegative for all sufficiently large $n$. Then
$$
\mathbb P(M_n-\ln n\le y)=\left(1-\frac{e^{-y}}n\right)^n\longrightarrow\boxed{\exp(-e^{-y})}.
$$
This is <convergence in distribution> to the standard <Gumbel distribution>.
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