(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.
Sketch of argument: For $x > 0$,
$$P(nZ_{(1)} > x) = P\left(Z_{(1)} > \frac xn\right)
= \prod_{k=1}^nP\left(Z_k > \frac xn\right) = \left(1-F_Z\left(\frac xn\right)\right)^n.$$
But for small $a>0$, $(1-a)^n \approx \exp(-na)$ (Taylor series match for first two terms) and so we have that
$$P(nZ_{(1)} > x) \approx \exp\left(-nF_Z\left(\frac xn\right)\right).$$ Now argue that $F_Z\left(\frac xn\right)$ is approximately equal to the limiting value $\lambda$ of the density $f_Z$ as the argument approaches $0$ times the length $\frac xn$ of the short interval $\left[0,\frac xn\right]$ over which we must integrate the pdf to find the CDF value at $\frac xn$ to arrive at the conclusion that
$$\lim_{n\to \infty} P(nZ_{(1)} > x) = \exp(-\lambda x),$$
and so the limiting distribution of $nZ_{(1)}$ is exponential with parameter $\lambda$. Putting in all the epsilons and deltas to make this a proof is left as an exercise for the OP.
Best Answer
The distribution function (CDF) of the minimum order statistic from an i.i.d. sample of $X$'s is
$$F_Z(z) = 1- [1- F_X(z)]^n$$
In our case $F_X(x) = 1-e^{-\lambda x}$ so
$$F_Z(z) = 1- [1- (1-e^{-\lambda z})]^n = 1-e^{-n\lambda z}$$
We are interested in the distribution of the random variable $Y\equiv n\lambda Z$.
A) The "CDF method" goes as follows:
$$P(Y \leq y) = P(n\lambda Z \leq y) = P\left( Z \leq \frac 1{n\lambda}y\right) = F_Z\left( \frac 1{n\lambda}y\right) = 1-e^{-n\lambda \frac 1{n\lambda}y}$$
$$\implies F_Y(y) = 1-e^{-y}$$
which is the CDF of an $\text{Exp}(1)$ random variable. In general, when the cumulative distribution function of the "source" random variable is closed formed/simple, this method is less prone to mistakes than the
B) "Change of Variable" method
The density function of $Z$ is
$$ f_Z(z) = n\lambda e^{-n\lambda z}$$
For the random variable $Y\equiv n\lambda Z$ we have $Z = \frac 1{n\lambda}Y$ and so
$$f_Y(y) = \left|\frac {\partial Z}{\partial Y}\right|\cdot f_Z\left(\frac 1{n\lambda}y\right) = \frac 1{n\lambda}\cdot n\lambda e^{-n\lambda \frac 1{n\lambda}y} = e^{-y}$$
which is the probability density function of an $\text{Exp}(1)$ random variable.