I want to prove that:
$$\lim_{n \rightarrow \infty} \sum_{k=1}^m {m \choose k} \Big( \frac{k}{m} \Big)^n = 1.$$
My progress so far: Define the function $F(x) \equiv (1+x)^m$ and the operator $\tilde{\nabla} \equiv x \cdot \frac{d}{dx}$. Using the binomial expansion on $F(x)$ it can be shown that:
$$\tilde{\nabla}^n F(x) \Big|_{x=1} = \sum_{k=1}^m {m \choose k} k^n \quad \quad \quad \text{for all } n \in \mathbb{N}.$$
So I effectively need to show that $\tilde{\nabla}^n F(x) \Big|_{x=1} \longrightarrow m^n$. I do not know how to express the left-hand-side in a useful closed form to do this.
Hint. The term for $k=m$ is just $1$. Now consider the sum of the remaining terms: given $m\geq 1$, $$0\leq \sum_{k=1}^{m-1} {m \choose k} \Big( \frac{k}{m} \Big)^n\leq \Big( \frac{m-1}{m} \Big)^n\sum_{k=1}^{m-1} {m \choose k}\leq \Big( \frac{m-1}{m} \Big)^n\cdot 2^m$$ because $1\leq k\leq m-1$ and $\sum_{k=1}^{m-1} {m \choose k}=2^m-1-1$.
What is the limit as $n\to \infty$ of the right-hand side?