Taken from Wikipedia:
The number $e$ is the limit $$e = \lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$
Graph of $f(x) = \left (1 + \dfrac{1}{x} \right)^x$ taken from here.
Its evident from the graph that the limit actually approaches $e$ as $x$ approaches $\infty$. So I tried approaching the value algebraically. My attempt:
$$\lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$ $$= \lim_{n \to \infty} \left(\frac{n + 1}{n}\right)^n$$ $$= \left(\lim_{n \to \infty} \left(\frac{n + 1}{n}\right) \right)^n$$ $$= 1^\infty$$
which is an indeterminate form. I cannot think of any other algebraic manipulation. My question is that how can I solve this limit algebraically?
An algebraic way is Binomial Expansion, which is given by $$\begin{eqnarray*} \left(1+\frac{1}{n}\right)^{\!n} &=& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^{\!2}+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^{\!3}+\cdots \\ \\ &=& 1+1+\frac{n(n-1)}{n^2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{n^3}\cdot \frac{1}{3!}+\cdots \\ \\ \end{eqnarray*}$$ Now every term in front there goes to $1$ for $n\to \infty$, so that the limit gives $$ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{\!n} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots = \mathrm{e}. $$