I am trying to show that for iid standard uniforms, $\displaystyle\lim _{ n\to \infty} P(X_{(1)} \le 1 + x/n) = e^{x} \mathbb I_{x \le 0}$, where $X_{(1)}$ is the maximum observation.
Using the fact that the distribution of maximums is :$F_{X_{(1)}}(x) = [F_X(x)]^n = x^n \mathbb I_{0\le x \le 1}$. We have that:
\begin{align*} \lim _{ n\to \infty} P(X_{(1)} \le 1 + x/n) &=\lim _{ n\to \infty}[( 1+ x/n)^n \mathbb I_{0 \le 1+ x/n \le 1}]\\ &=\lim _{ n\to \infty}[( 1+ x/n)^n \mathbb I_{-n \le x \le 0}]\\ &= e^x \mathbb I_{-\infty \le x \le 0}\\ &= e^x \mathbb I_{x \le 0} \end{align*}
Intuitively the last step makes sense, I'm just wondering if this is rigorous in the sense that going from the limit of the products to just plugging in $-n = -\infty$?
First of all, a minor technicality regarding your statement: For $x>0$ both the LHS and RHS of your statement should be equal to 1 (while with your indicator notation the RHS would be 0, while the LHS is still 1)! What you really want to show is the following:
$$\displaystyle\lim _{ n\to \infty} P(X_{(1)} \le 1 + x/n) = e^{x} \; \forall \; x\leq0$$
If you write it out like that, it might already clear up your confusion. But to answer, yes, your proof is fine. To be a bit more formal you could apply the product rule for limits from the second to the third line of the proof.
Also, if it makes you feel more comfortable with the derivation, you could also proceed as follows: Fix $x < 0$ (you are doing this in your post implicitly anyway, the case $x$=0 is easy). Then because $1 + \frac{x}{n} \to 1$ as $n\to\infty$, there exists $n_0$ such that $1 + \frac{x}{n} \geq 0 \; \forall n \; \geq n_0$. OTOH it also holds that $1 + \frac{x}{n} \leq 1$. Therefore just write out everything for $n \geq n_0$ and then take $n\to\infty$, by doing this you can basically ignore the indicator functions in the derivation.