Prove that $\sin(ne^{-nx}), x\in E=[1;+\infty)$ converges uniformly.

Question

Here is a function: $f_n(x)=\sin(ne^{-nx}), x\in E=[1;+\infty)$. And I have to prove its uniform convergence.
It is obvious that $f_n(x)\rightarrow f(x)=0$ when $n\rightarrow \infty$ since $\lim_{n\rightarrow \infty}\sin\frac{n}{e^{nx}}=\sin0=0$.
The problem is to prove that: $$ \exists \{a_n\}, \exists n_0: \forall n\geqslant n_0, \forall x\in E\ \ |f_n(x)-f(x)|<a_n,\ \lim_{n\rightarrow\infty}a_n=0 $$ I just cannot find the right $a_n(n)$ function.

Answer 1

$$|\sin(ne^{-nx})| \leq ne^{-nx} \leq ne^{-n}, \forall x \geq 1$$

Answer 2

$|\sin x| \leq x$ for all $x \geq 0$ and $0\leq ne^{-nx}\leq ne^{-n}$. Can you show that $ne^{-n} \to 0$ to conclude that the given functions converge uniformly to $0$?