(The answer has been reworked to respond to OP's and whuber's comments).
The complementary cdf of $X$ is
$$G_n(x) = \left[1-F_Z\left(x/n\right)\right]^{n}$$
To prove that asymptotically $X$ follows an exponential distribution, we need to show that $$\lim_{n\rightarrow \infty}G_n(x)= e^{-\lambda x}$$
Consider
$$F_Z\left(x/n\right) = \int_0^{x/n}f(t)dt $$
By the properties of the integral, we have
$$\int_0^{x/n}f(t)dt = \frac 1n\int_0^{x}f(t/n)dt$$
Define
$$h_n(w) = \left(1+\frac {w}{n}\right)^{n}, \qquad \lim_{n\rightarrow \infty}h_n(w) = e^w=h(w), \;\; w \in \mathbb R$$
and
$$g_n(x) = -\int_0^{x}f(t/n)dt,\;\;\; -\lim_{n\rightarrow \infty}g_n(x) = -\int_0^{x}f(0)dt = -\lambda x = g(x), \;\;x \in \mathbb R_+$$
(To respond to a question by the OP, we can take the limit inside the integral. First note that $n\geq 1$, and we do not send $x$ to infinity. So the argument of $f$ does not explode. So even if it were the case that $f(\infty) \rightarrow \infty$, we do not need to consider this case here. Then, since also $f(0)$ is finite by assumption, $f$ is bounded and dominated convergence holds).
With these definitions we can write
$$G_n(x) = h_n(g_n(x))$$
and the question is
$$ \lim_{n\rightarrow \infty}h_n(g_n(x)) =?\;\; h(g(x)) = e^{-\lambda x},\;\;x \in \mathbb R_+$$
The limit of a composition of function-sequences does not in general equal the composition of their limits (which is what whuber has essentially pointed out in his comment). But this equality will hold if
$(i)$ $h_n$ converges uniformly to $h$ (it does-convergence to $e^w$ is uniform)
$(ii)$ the limit of $h_n$ is a continuous function (it is)
$(iii)$ the functions $g_n(x)$ map $\mathbb R_+$ to $\mathbb R$ (namely, they map their domain into the set where $h_n$ converges -they do).
So the above equality holds and we have proven what we needed to prove.
I'm going to assume this is self-study, and make some suggestions rather than giving an explicit answer. The proof is in two steps:
- Find the joint distribution function of $U_{(1)}, U_{(n)}$;
- Show that in the limit as $n \to \infty$ the joint CDF factors.
There is an alternative: you could find the joint characteristic function (or moment generating function, if you prefer) and again show that the limiting characteristic function factors.
Best Answer
Taking your CDF and using whuber's hint, we have the expectation of the maximum as $$E[X_{(n)}] = \int_0^\lambda \left[ 1 - \left( {x \over \lambda} \right)^n \right]dx=\lambda - {\lambda \over {n+1}} $$ This simplifies to $$E[X_{(n)}]= \lambda \left[ {n \over {n+1}} \right]. $$ Then for your altered estimator we get $$E[\lambda^*]=\left[ {{n+1} \over n} \right] E[X_{(n)}]= \lambda.$$