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.
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$\begingroup$I'd do it like this:\begin{align} \left(1+\frac{x}{n}\right)^{n}&=\sum_{k=0}^{n}{\begin{pmatrix}n\\k\end{pmatrix}1^{n-k}\left(\frac{x}{n}\right)^{k}}&\\ &=\sum_{k=0}^{n}{\frac{n!}{k!\;(n-k)!}\left(\frac{x}{n}\right)^{k}}&\\ &=\sum_{k=0}^{n}{\frac{(n-k)!\;(n-(k-1))\ldots n}{k!\;(n-k)!}1^{n-k}\left(\frac{x}{n}\right)^{k}}&\\ &=\sum_{k=0}^{n}{\frac{(n-(k-1))\ldots n}{k!}\frac{x^{k}}{n^{k}}}&\\ &=\sum_{k=0}^{n}{\frac{x^{k}}{k!}\frac{n-(k-1)}{n}\ldots\frac{n}{n}}&\\ &=\sum_{k=0}^{n}{\frac{x^{k}}{k!}\left(1-\frac{k-1}{n}\right)\ldots}& \end{align}Then apply the limit:\begin{align} &\lim_{n\rightarrow\infty}{\sum_{k=0}^{n}{\frac{x^{k}}{k!}\left(1-\frac{k-1}{n}\right)\ldots}}&\\ &=\sum_{k=0}^{\infty}{\frac{x^{k}}{k!}}& \end{align}Ta-dah. Hope this helps.
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