Convergence of $\left(1+\frac{1}{n}\right)^n$, given that $\left(1+\frac{1}{2^n}\right)^{2^n}$ converges I know how to prove that the sequence
$$\left(1+\frac{1}{2^n}\right)^{2^n}$$ 
has a limit.
Can I use this knowledge to quickly get the fact that 
$$\left(1+\frac{1}{n}\right)^n$$ 
also has a limit?
 A: First notice that if $a_n$ is an increasing sequence, then $a_n$ converges iff $a_{2^n}$ converges. The forward implication is true since every subsequence of a convergent sequence converges. The reverse implication follows since for every $m \in \mathbb{N}$ there is an $n \in \mathbb{N}$ for which $$a_{2^n} \le a_m < a_{2^{n+1}}$$ then convergence is obtained via the squeeze theorem.
Now demonstrate that the sequence $$\left( 1+\frac1n\right)^n$$ is increasing. You may do this by writing $$f(x) = \left( 1+ \frac1x\right)^x$$ and by logarithmic differentiation you can show: $$f'(x) = \left(1+\frac1x\right)^x \left( \ln\left( 1+ \frac1x \right) + \frac{x^2}{1+x} \right)$$ which is positive for positive $x$. Hence $f(n+1) > f(n)$.
Hence $$\left( 1 + \frac1n\right)^n$$ is an increasing sequence. Now since you claim you can prove $$\left( 1 + \frac1{2^n}\right)^{2^n}$$ converges, you may conclude that the original sequence converges.
A: You are trying to say that since $f(2^n)$ has a limit (as $n$ goes to infinity), then also $f(n)$ has a limit as $n$ goes to infinity. This is not in general true. Take for example
$$
f(n) = \sin(2\pi\log_2(n)).
$$
Then
$$
\lim_{n\to \infty} f(2^n)= \lim_{n\to \infty} \sin(2\pi n) = 0.
$$
But
$$
\lim_{n\to \infty} f(n)
$$
does not exist.

This might be helpful: https://math.stackexchange.com/a/580262/26188
A: In case you are confused with this result. 
$$\lim\limits_{x\rightarrow p}{f(x)}=q$$
if and only if 
$$\lim\limits_{n\rightarrow\infty}{f(p_n)}=q$$
for every sequence $\{p_n\}$ such that $$p_n\neq p, \lim\limits_{n\rightarrow \infty}{p_n}=p$$
A: This is just a follow-up to the very nice answer by Joel and the discussion in comments by gniourf_gniourf.  
To show that $(1+{1\over n})^n$ is increasing, it suffices to show that $(1+{1\over n})^n/(1+{1\over n+1})^{n+1}\lt1$.  
Some preliminary algebra gives
$$\begin{align}
{\left(1+{1\over n}\right)^n\over\left(1+{1\over n+1}\right)^{n+1}}&={\left({n+1\over n}\right)^n\over\left({n+2\over n+1}\right)^{n+1}}\\
&=\left({n+1\over n+2}\right)\left({(n+1)^2\over n(n+2)}\right)^n\\
&=\left({n+1\over n+2}\right)\left({n^2+2n+1\over n^2+2n}\right)^n\\
&=\left({n+1\over n+2}\right)\left(1+{1\over n(n+2)}\right)^n\\
\end{align}$$
The binomial expansion and some crude bounds give
$$\begin{align}
\left(1+{1\over n(n+2)}\right)^n
&=1+{n\choose1}{1\over n(n+2)}+{n\choose2}{1\over n^2(n+2)^2}+\cdots\\
&=1+{1\over n+2}+{n(n-1)\over2\cdot1\cdot n^2}{1\over(n+2)^2}+{n(n-1)(n-2)\over3\cdot2\cdot1\cdot n^3}{1\over(n+2)^3}+\cdots\\
&\lt1+{1\over n+2}+{1\over2}{1\over(n+2)^2}+{1\over3\cdot2}{1\over(n+2)^3}+{1\over4\cdot3\cdot2}{1\over(n+2)^4}+\cdots\\
&\lt1+{1\over n+2}+{1\over2}{1\over(n+2)^2}+{1\over2^2}{1\over(n+2)^3}+{1\over2^3}{1\over(n+2)^4}+\cdots\\
&=1+{1\over n+2}\left(1+{1\over2n+4}+{1\over(2n+4)^2}+{1\over(2n+4)^3}+\cdots \right)\\
&=1+{1\over n+2}\left({1\over1-{1\over2n+4}} \right)\\
&=1+{1\over n+2}\left({2n+4\over2n+3} \right)\\
&=1+{2\over2n+3}\\
&={2n+5\over2n+3}
\end{align}$$
Putting this into the previous equality gives
$${\left(1+{1\over n}\right)^n\over\left(1+{1\over n+1}\right)^{n+1}}
\lt{n+1\over n+2}\cdot{2n+5\over2n+3}={2n^2+7n+5\over2n^2+7n+6}=1-{1\over2n^2+7n+6}\lt1$$
as requested.  I've included a lot of the intermediary algebraic steps to make the proof easy to follow from line to line.  Aside from abridging the presentation, I don't see any obvious way to shorten the proof.  I'd be happy to see something shorter.
