TITLE: Do the polynomials $(1+z/n)^n$ converge compactly to $e^z$ on $\mathbb{C}$?
QUESTION [6 upvotes]: The question is

Do the polynomials $p_n(x)=(1+z/n)^n$ converge compactly (or uniformly on compact subsets) to $e^z$ on $\mathbb{C}$?

I thought about expanding
$$p_n(z)=\sum_{k=0}^n a_k^{(n)}z^k$$
where
$$a_k^{(n)}=\binom{n}{k}\frac{1}{n^k}=\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)$$
and trying to show that $\frac{1}{k!}-a_k^{(n)}$ decreases sufficiently fast on any closed ball. That is, I tried to show
$$\lim_{n\rightarrow\infty}\max_{z\in\overline{B_0(A)}}\left|\sum_{k=0}^n\frac{z^k}{k!}-p_n(z)\right|=0$$
for any fixed $A>0$, but I had difficulty with this approach.
Any help is appreciated.

REPLY [2 votes]: Here's a less computational way to get the result.
Lemma 1: If $f_n: E \to \mathbb C $ are uniformly bounded and converge uniformly to $f$ on $E,$ then $e^{f_n} \to e^f$ uniformly on $E.$ Proof: Easy.
Lemma 2: Let $\log $ denote the principal value logarithm. Then there is a constant $C$ such that for $0<|u|< 1/2,$ 
$$\left|\frac{\log (1+u)}{u}-1\right| \le C|u|.$$
Proof: Inside the absolute values on the left we have a function that extends to be analytic in $D(0,1)$ with value $0$ at $0.$ The result follows easily.
So let $R>0.$ For $0<|z|<R$ and $n>2R,$ Lemma 2 shows
$$|n\log(1+z/n) - z| = |z|\,\left|\frac{\log(1+z/n)}{z/n}-1\right| \le 
R\cdot C(R/n) = CR^2/n.$$
So we have uniform convergence of $n\log(1+z/n)$ to $z$ on $D(0,R),$ at least along the sequence of $n$'s greater than $2R.$ Because these functions are uniformly bounded on $D(0,R),$ Lemma 1 shows
$$\tag 1 \exp (n\log(1+z/n)) \to e^z$$
uniformly on $D(0,R).$ Since the left side of $(1)$ equals $[\exp (\log(1+z/n))]^n = p_n(z),$ we're done.