Showing the Bernstein Polynomial for $e^x$ converges I am trying to show that the Bernstein polynomial for $f(x)=e^x$ on $[0,1]$ converges uniformly to $f$. 
I've shown that the nth Bernstein polynomial for $f$ is: $$B_n(f,x)=(1-x+xe^{\frac{1}{n}})^n.$$
So I'm left to show that:
$$\forall{\space{}}\epsilon>0\space{}\text{ there exists an }N\in{\mathbb{N}}\text{ s.t. }(n\ge{N})\implies(|B_n(f,x)-f(x)|<\epsilon)\text{ for all }x\in{[0,1]}.$$
Any hints will be appreciated.
 A: This holds in general:

If $f\in C[0,1]$, then the sequence of Bernstein polynomials $B_nf$ converges uniformly to $f$ on $[0,1]$.

Its proof is pretty involved: see Ross, Elementary Analysis, theorem $27.4$.
A: The first important thing to see is that $n(1-e^{1/n})\to1$, as $n\to\infty$, which you can show in any way you like. Now the following lemma instantly gives you the result you are looking for:

Lemma. Let $a_n$ be a sequence converging to $a$. Then the functions
  $$ f_n : \Bbb R\to\Bbb R, \quad x\mapsto \left(1+ \frac{a_n x}{n}\right)^n$$
  converge uniformly on compacta to the function $x\mapsto e^{ax}$.

To prove the lemma we will show uniform convergence on a set of the form $[-R,R]$. Now consider $x\in[-R,R]$:
$$\left|\left(1+ \frac{a_n x}n\right)^n - e^{a x}\right| ≤ \sum_{k=0}^N \left| \frac{n!}{(n-k)!n^k} \,a_n^k - a^k\  \right| \frac{R^k}{k!} + \sum_{k=N+1}^n \left|\frac{n!}{(n-k)!} \frac{a_n^k}{n^k}\right|\frac{R^k}{k!} +\sum_{k=N+1}^\infty  a^k\frac{R^k}{k!}.$$
Now we just understand these terms and get the reuslt. Note that for any fixed $N$ the first summand goes to $0$ as $n\to\infty$. For the second summand notice that the $\frac{n!}{(n-k)!} \frac{a_n^k}{n^k}$ term will for $n$ large enough be smaller than $(|a|+1)^k$, hence this summand can be bounded by $\sum_{k=N+1}^\infty \frac{(|a|+1)^k R^k}{k!}$, which can be made as small as you please for $N$ large enough. Similarly the third summand can be made as small as you like, you just need to choose $N$ large enough.
This retrieves a bound for the difference that you can make as small as you want, and this can be done independently of $x$ (provided $|x|≤R$), giving you the desired result.
