$e\approx2.7182818284\dots$ is an irrational number. It has several representations. I'll give a few representation, along with a slight generalization.
\begin{align}
e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\\
e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n
\end{align}
This is the usual definition. The $\lim$ sign means, roughly, that you let $n$ get larger and larger. For example, $(1+\frac1{1000})^{1000}=2.7169\dots\approx e$, and you get better and better approximations as $n$ gets bigger. To prove the second thing from the first one, replace $n$ with $\frac nx$, and notice that $\frac nx\to\infty$ means the same thing as $n\to\infty$ (when $x$ is positive, at least. It works for negative $x$, too, though).
\begin{align}
e&=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\dotsb\\
e^x&=\ 1\ +\ x\ +\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb
\end{align}
Remember that $0!=1$. You can try to prove this from the first set of equalities and the binomial formula. That's how Euler did it. (If you want to be rigorous, though, you need more effort. Euler was never very rigorous in this sort of thing.)
\begin{align}
\int_1^e\frac1x\operatorname d\!x&=1\\
\int_1^t\frac1x\operatorname d\!x&=\log_e(t)
\end{align}
I need to explain the notation. The shape surrounded by the curve $y=\frac1x$, the vertical lines $x=1$ and $x=e$, and the $x$-axis has area $1$. More generally, the shape bounded by $\frac1x$, $x=1$, $x=t$, and the $x$-axis has area $\log_e(t)$. I won't prove this here.
And my favorite characterization: $e$ is the unique number that satisfies:
$$e^x\ge x+1$$
for all $x$.