The proof will be given by demonstrating that the sequence (1) is:

1. 1.

   monotonic (increasing), that is $a_{{n}}<a_{{n+1}}$

2. 2.

   bounded above, that is $\forall n\in\mathbb{N},a_{n}<M$ for some $M>0$

In order to prove part 1, consider the binomial expansion for $a_{n}$:

Since $\forall i\in\{1,2\ldots(k-1)\}:(1-\frac{i}{n})<(1-\frac{i}{n+1})$, and since the sum $a_{{n+1}}$ has one term more than $a_{n}$, it is demonstrated that the sequence (1) is monotonic.
In order to prove part 2, consider again the binomial expansion:

Since $\forall k\in\{2,3\ldots n\}:\frac{1}{k!}<\frac{1}{2^{{k-1}}}$ and $\frac{n(n-1)\ldots(n-(k-1))}{n^{k}}<1$:

where the formula giving the sum of the geometric progression with ratio $1/2$ has been used. ∎