<s>[INST] Direct proof that $\sum_{n\geq 0}\frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$ Is there a direct proof that $$\sum_{n\geq 0}\frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n?$$
We dont know what logarithms or exponentials are.
 [/INST] Applying the Binomial theorem,
\begin{equation*}
\lim_{n\to\infty}(1+\frac{x}{n})^n = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\frac{(n-2)(n-1)x^3}{6n^2}+\frac{(n-3)(n-2)(n-1)x^4}{24n^3}+\frac{(n-4)(n-3)(n-2)(n-1)x^5}{120n^4}+... \\
=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+... \\
=\sum_{n\geq 0}\frac{x^n}{n!}.
\end{equation*}
Note that we used L'Hopital's rule when taking the limits because we had the form $\frac{\infty}{\infty}$. 
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