If $\lim_{n \to \infty} n x_n = 0$ then $\lim_{n\to \infty} (1+x_n)^n =1$

The following appears without proof in some lecture notes about exponents (specifically, it was used to prove the limit $\lim_{n\to \infty} (1+\frac{x}{n})^n = e^x := \sup\{ e^q \mid q\in \Bbb Q, q\le x\}$ for all $x\in \Bbb R$):

Proposition. If $\lim_{n \to \infty} n x_n = 0$ and $x_n \ge 0$ for all $n\in \Bbb N$, then $\lim_{n\to \infty} (1+x_n)^n =1$.

I tried to give a proof, for the sake of completeness:

Proof. We have $0<\frac{1}{1+x_n} \le 1$ for all $n$, so we can write $\frac{1}{1+x_n} = 1-h_n$ for some nonegative $0\le h_n < 1$. That is, let $h_n = \frac{x_n}{x_n+1}$, and obviously $\lim_{n\to \infty} h_n = 0$ (since $\lim_{n\to \infty} x_n = \lim_{n\to \infty} \frac{h_n}{h_n-1}= 0$). Now,$$ \lim_{n \to\infty} nx_n = \lim_{n \to\infty} \frac{nh_n}{h_n-1} = 0\ \ \Longrightarrow \ \ \lim_{n\to \infty} nh_n =0.$$

So, from Bernoulli's inequality,

$$1\ge \frac{1}{(1+x_n)^n} = (1-h_n)^n\ge1-nh_n.\ \ \ \ \ $$The rest follows from the squeeze theorem.$$\tag*{$\square$}$$

Is this proof correct? I'm surprised I didn't find the answer on this site, but I might have missed a similar question.