We have $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} $$ and $$\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^n \frac1{n^k}{n\choose k}x^k $$ therefore $$\left\lvert e^x-\left(1+\frac{x}{n}\right)^n\right\rvert\le\sum_{k=n+1}^\infty \frac{\lvert x\rvert^k}{k!}+\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert. $$
Question: As $n\to\infty$, the first sum trivially goes to $0$, but how to properly bound the second one?
I can show convergence in different ways but I'm interested in this specific approach.