Characterization of the number $e$ The number $e$ is characterized in calculus as: 
$$e:=\lim_{n\to\infty}\left(1+\frac1n\right)^n$$
My question is this:
As $n$ heads to infinity, $1/n$ gets closer and closer to zero. So what we're really doing is to add $1$ to a number that's getting closer and closer to zero and then raising that expression to an increasingly large number. 
So, the limit as $n$ moves towards infinity is $1$. Why is it $e$?
 A: Here is a table of the sequence $\left(1 + \frac{1}{n}\right)^n$ from Wolfram Alpha:

You can see the sequence starts at $2$ and increases.  So its limit cannot be $1$.
You write:

So what we're really doing is to add 1 to a number that's getting closer and closer to zero and then raising that expression to an increasingly large number.

This is basically what you're saying:
$$
\color{red}{
\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n
= \lim_{n\to\infty} 1^n
= 1}
\qquad\text{(false)}
$$
But in this argument, you split the $n$ in the base and the $n$ in the exponent, and took the limit as they tended to $\infty$ separately.  But that's not allowed; it's the same variable in both, so we can only take the limit once.
A: Let $N$ be a fixed integer. Then as $n$ goes to infinity


*

*on one hand,
$$\left(1+\frac1n\right)^N\to1$$

*and on the other
$$\left(1+\frac1N\right)^n\to\infty.$$


You are focusing on the "1." side, but the "2." side gives a completely different conclusion. In fact, there is no fixed $N$ and all you can tentatively infer for now is that the limit lies somewhere between $1$ and $\infty$ (if it exists). 

Using the binomial expansion of $\left(1+\dfrac1n\right)^n$, one can show that the actual limit is given by the converging series
$$e:=\sum_{k=0}^\infty\frac1{k!}.$$
A: I though similarly, but the reason is because even though you think you are adding just $1$, this is not the case. It is true that the second term $\text{approaches 0}$, but it never quite gets there. Thus you always have $1 + \text{some small number}$, raised to exponent, and thus the limit goes to $e$ if you look at Matthew's chart above. It is the $\frac{1}{n}$ that curves the graph upwards, because you are still $\text{adding}$ by a small amount. Although this amount gets less and less and $n \to \infty$, it never disappears.
A: As @MyGlasses comments, a quick binomial expansion reveals that
$$\left(1+\frac1x\right)^x=1+1+\frac12\frac{x-1}x+\dots\\=2+\frac12\frac{x-1}x+\dots\\>2$$
And from experience, you should realize that all terms in the binomial expansion will be positive.
