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?

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.
