= Solution
<Varadhan's lemma> states that if $U_n$ has a <large deviation principle> with <good rate function> $K$ and <large-deviation speed> $a_n$, then for every bounded <continuous> real function $h$,
$$
\lim_n\frac1{a_n}\log\mathbb E e^{a_nh(U_n)}=\sup_u\{h(u)-K(u)\}.
$$
For a continuous function unbounded above, the same conclusion holds under the <exponential tail extension of Varadhan's lemma>, for example when the weighted contribution from $h>M$ is superexponentially negligible as $M\to\infty$.
The <scalar clipping> map $g(m)=\min(b,\max(a,m))$ is continuous. The <rate function under scalar clipping> has endpoint costs $\inf_{m\le a}J(m)$ and $\inf_{m\ge b}J(m)$. Since $a<\mu<b$ and the costs decrease towards $\mu$ from the left and increase to its right, both infima are attained at the corresponding endpoints. Thus the <good rate function> of $Z_n$ equals $J(z)$ on $[a,b]$ and is infinite outside. The function $\log z$ is bounded and continuous on that compact interval, so
$$
\lim_n\frac1n\log\mathbb E[Z_n^n]
=\max_{a\le z\le b}\{\log z-J(z)\}.
$$
On $[a,\mu]$, the derivative of $\log z-I(z)$ is $2/z-\lambda>0$, so the maximum there is $\log\mu$. On $[\mu,b]$, the derivative is $(k+1)/z-k\lambda$, with negative second derivative. Its unconstrained maximizer is
$$
z_* =\frac{k+1}{k\lambda}.
$$
It lies in $[\mu,b]$ whenever $k\ge(\lambda b-1)^{-1}$. The right-hand objective increases initially at $\mu$, so this maximizer dominates the whole left-hand interval. Substitution gives the limit for <high moments of a clipped minimum of exponential sample means>
$$
\boxed{\lim_n\frac1n\log\mathbb E[Z_n^n]
=(k+1)\log\frac{k+1}{k}-\log\lambda-1.}
$$
The moment here is the <expectation> of the $n$th power. If the printed notation were instead read as $(\mathbb E Z_n)^n$, its logarithmic limit would be $\log\mu=-\log\lambda$, since boundedness and <convergence in probability> give $\mathbb E Z_n\to\mu$.
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