real analysis - Uniform convergence of $f_n(x) = left(1 + frac{x}{n}right)^n$ when calculating limit

Calculate $$\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac{x}{n}\right)^ndx$$

My attempt - if
$$f_n(x) = \left(1 + \frac{x}{n}\right)^n$$
converged uniformly for all $x \in [0,1]$ then I could swap integral with limes and solve it:
$$\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac{x}{n}\right)^ndx = \int_0^1 \lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^ndx = \int_0^1 e^x dx = e^x|_{0}^{1} = e - 1$$

I must then prove that $f_n(x)$ is indeed uniformly convergent. I already know that

$f_n(x) \rightarrow e^x$. If $f_n(x)$ converges uniformly then for each epsilon the following statement must hold
$$\sup_{x \in [0,1]} \left|f_n(x) - f(x)\right| < \epsilon$$

How can I prove this?

Answer

Alternative approach (without uniform convergence): let $t= \frac{x}{n}$ then
$$\begin{align*}\int_0^1 \left(1 + \frac{x}{n}\right)^ndx&= n\int_0^{1/n} (1+t)^ndt\\ &=n\left[\frac{(1+t)^{n+1}}{n+1}\right]_0^{1/n} \\&= \frac{n}{n+1}\left(\left(1+\frac{1}{n}\right)^{n+1}-1\right)\\&\to e-1. \end{align*}$$