Need to prove the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving it is a Cauchy sequence 
I am trying to prove that the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving that it is a Cauchy sequence. 

I don't get very far, see: for $\epsilon>0$ there must exist $N$ such that $|a_m-a_n|<\epsilon$, for $ m,n>N$
$$|a_m-a_n|=\bigg|\bigg(1+\frac{1}{m}\bigg)^m-\bigg(1+\frac{1}{n}\bigg)^n\bigg|\leq \bigg|\bigg(1+\frac{1}{m}\bigg)^m\bigg|+\bigg|\bigg(1+\frac{1}{n}\bigg)^n\bigg|\leq\bigg(1+\frac{1}{m}\bigg)^m+\bigg(1+\frac{1}{n}\bigg)^n\leq \quad?$$
I know I am supposed to keep going, but I just can't figure out the next step. Can anyone offer me a hint please? Or if there is another question that has been answered (I couldn't find any) I would gladly look at it.
Thanks so much!

Look if it works!
For $m>n \geq 1, |a_m-a_n| \leq |a_m^{1/m}-a_n^{1/n}|=|\frac {1}{m}-\frac {1}{n}|<\frac {1}{m}+\frac {1}{n}<\frac {2}{n}< \epsilon$ if $n >[2 /\epsilon].$
