Why does this limit approach $e$? Why does
 $$(1+1/k)^{k}\rightarrow e\hspace{0.5cm}\text{as}\hspace{0.5cm}k\rightarrow\infty$$
why does it not approach 
$$(1+1/k)^{k}\rightarrow (1+0)^k=1\hspace{0.5cm}\text{as}\hspace{0.5cm}k\rightarrow\infty$$
 A: What you are correctly observing is:
$$\lim_{m \to \infty} \lim_{n \to \infty} (1+1/n)^m = \lim_{m \to \infty} 1 = 1.$$
To quickly see why something weird might happen, notice that
$$\lim_{n \to \infty} \lim_{m \to \infty} (1+1/n)^m = \lim_{n \to \infty} \infty = \infty.$$
In these two cases, we send $n,m$ to infinity separately and consecutively. In your limit, you send $n,m$ to infinity simultaneously and in fact at the same rate. Thus you are raising a number that is getting closer and closer to $1$ to a larger and larger power, and the two effects compete with each other to give a limit which is neither $1$ nor $\infty$.
A: The reason is that $1 + 1/k$ is decreasing to $1$ while the exponent $k$ increases to $+\infty$, resulting in the indeterminate form $1^\infty$.
The expression $\lim_{k \to \infty} (1+1/k)^k$ is the definition of $e$.
A: Interpreting the question as why the limit is $>1$, the answer is that the sequence is strictly increasing. See The Limit Definition of e.
A: By the Taylor series expansion, as $x \to 0$, one obtains
$$
\ln(1+x)=x+x\epsilon(x)
$$ with $\displaystyle \epsilon(x) \to 0$, as $x \to 0$, giving as $n \to \infty$,
$$
n\ln\left(1+\frac1n\right)=1+\epsilon\left(\frac1n\right) \to 1
$$ that is
$$
\left(1+\frac1n\right)^n=e^{n\ln\left(1+\frac1n\right)} \to e^1=e.
$$
