= Solution
The <exponential distribution> has no atoms, so all sample values are distinct <almost surely>, and $M_i$ is the $i$th <order statistic>. Set $M_0=0$ and $D_i=M_i-M_{i-1}$. Part (b) gives $D_1\sim\operatorname{Exp}(n\lambda)$.
To justify induction and the required <independence>, let $J$ be the label attaining the minimum. Conditional on its time and label, the remaining clocks are fresh <independent> rate-$\lambda$ <exponential random variables>, by the <memoryless property>. This conditional claim can be checked directly: for a specified label $j$, first ringing time $u$, and residual lifetimes $y_k>0$ for $k\ne j$, the <joint probability density> is
$$
\lambda^n e^{-\lambda(nu+\sum_{k\ne j}y_k)}
=\frac1n\left(n\lambda e^{-n\lambda u}\right)
\prod_{k\ne j}\left(\lambda e^{-\lambda y_k}\right).
$$
Thus the first minimum, its uniform winning label and the surviving residual clocks are <independent>. Repeat this argument after each minimum. At the $i$th step there are $n-i+1$ clocks, so the <independent exponential order-statistic spacings> satisfy
$$
D_i\sim\operatorname{Exp}((n-i+1)\lambda),\qquad D_1,\ldots,D_n\text{ independent}.
$$
Since $Y_k/k$ has rate $k\lambda$ whenever $Y_k\sim\operatorname{Exp}(\lambda)$, summing the gaps proves
$$
\boxed{M_i\overset d=\sum_{k=n-i+1}^n\frac{Y_k}{k}.}
$$
In particular the maximum is $M^*=M_n$. By <variance additivity for independent random variables> and the <Laplace transform of an exponential distribution>, its requested quantities are
$$
\boxed{\mathbb EM^*=\frac1\lambda\sum_{k=1}^n\frac1k,\qquad
\operatorname{Var}(M^*)=\frac1{\lambda^2}\sum_{k=1}^n\frac1{k^2},\qquad
\mathbb E e^{-sM^*}=\prod_{k=1}^n\frac{k\lambda}{k\lambda+s}\quad(s\geq0).}
$$
The product follows from <independence> of the gaps, not merely from their marginal <probability distributions>.
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