I try to prove the Exponential series :
$$\exp(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k!}$$
From the definition of the exponential function $$\exp(x) \stackrel{\mathrm{def}}{=} \lim_{n\to\infty} \left(1+\dfrac{x}{n}\right)^n$$
I've tried a Binomial expansion of $\exp(x)$ like : $$\begin{split} \exp(x) &= \lim_{n\to\infty} \sum_{k=0}^{n} \binom{n}{k}\dfrac{x^k}{n^k}\\ &= 1 + \lim_{n\to\infty} \sum_{k=1}^{n}\left(\dfrac{x^k}{k!}\times \dfrac{n!}{(n-k)!\times n^k}\right)\\ &= 1 + \lim_{n\to\infty} \sum_{k=1}^{n}\dfrac{x^k}{k!}\prod_{j=1}^{k}\left(\dfrac{n-(j-1)}{n}\right)\\ &= 1 + \lim_{n\to\infty} \sum_{k=1}^{n}\dfrac{x^k}{k!}\prod_{j=1}^{k}\left(1-\dfrac{j-1}{n}\right)\\ \end{split}$$
Here is my problem. If I apply the limit, obtain : $$\lim_{n\to\infty} \dfrac{j-1}{n} = (j-1) \times \lim_{n\to\infty}\dfrac{1}{n} = 0$$
But $j$ approaches $k$ which approaches $n$, so $j$ approaches the infinity... and the limit is indeterminate : $\infty \times 0 = \,?$
How to evaluate this indeterminate form?
Thanks in advance.