Importance of Tannery's theorem 
Tannery's theorem:
Let $S_n=\sum_{k=0}^\infty a_k(n)$ and $\lim_{n\to\infty}a_k(n)=b_k$. If $|a_k(n)|\le M_k$ and $\sum_{k=0}^\infty M_k\lt\infty$, then $\lim_{n\to\infty}S_n=\sum_{k=0}^\infty b_k$.

This can be used to prove that
$$\lim_{n\to\infty}\sum_{k=0}^n \frac{x^k}{k!}\prod_{m=1}^{k-1}\left(1-\frac{m}{n}\right) =\sum_{k=0}^\infty \frac{x^k}{k!}.$$
But I've never really understood why is the theorem "needed", as $\prod_{m=1}^{k-1}\left(1-\frac{m}{n}\right)$ tends to $1$ with increasing $n$, so the result should be obvious.
What could go wrong here? Could someone provide a counterexample to this reasoning?
 A: This is just a special case of a combination of (1) the Weierstrass M-test for uniform convergence and (2) the result that the limit can be exchanged with the sum when convergence is uniform.
(1) Suppose we have a sequence of functions $x \mapsto a_k(x)$ where $x \in D \subset \mathbb{R}$ and $|a_k(x)| \leqslant M_k$ for all $k$ and for all $x \in D$.  If  $\sum_{k=0}^\infty M_k < \infty$, then the series $\sum a_k(x)$ is unformly convergent for $x \in D$.
(2) If $x_0 \in D$, then under the conditions of (1) where $\sum a_k(x)$ is uniformly convergent, we have $\lim_{x \to x_0} \sum_{k=0}^\infty a_k(x) =  \sum_{k=0}^\infty \lim_{x \to x_0} a_k(x)$. This holds when $x_0 = +\infty$ as well.
Tannery's theorem can be seen to be a special case of these results by taking $D = \mathbb{N}$ and noting that
$$S_n = \sum_{k=0}^na_k(n) = \sum_{k=0}^\infty a_k(n) \mathbf{1}_{\{k \leqslant n\}},$$
since $|a_k(n)| \leqslant M_k$ implies that $ |a_k(n) \mathbf{1}_{\{k \leqslant n\}}| \leqslant M_k$.
What can go wrong is that without sufficient conditions as in uniform / monotone / dominated convergence it is not always possible to exchange the limit and the sum.
