Find the last two digits of $ 7^{81} ?$ I came across the following problem and do not know how to tackle it.  

Find the last two digits of $ 7^{81} ?$  

Can someone point me in the right direction? Thanks in advance for your time.

$\rm{\bf Hint}\  \ mod\,\ \color{#c00}2n\!: \ a\equiv b\,  \Rightarrow\,  mod\,\ \color{#c00}4n\!:\ \, a^2 \equiv b^2\ \  by \ \ a^2\! = (b\!+\!\color{#c00}2nk)^2\!=b^2\!+\!\color{#c00}4nk(b\!+\!nk)\equiv b^2$   
$\rm So,\,  \ mod\,\ \color{}50\!:\, 7^{\large 2}\!\equiv -1\Rightarrow mod\ \color{}100\!:\,\color{#0a0}{7^{\large 4} \equiv\, 1}\:\Rightarrow\:7^{\large 1+4n}\equiv\, 7\, \color{#0a0}{(7^{\large 4})}^{\large n} \equiv\, 7 \color{#0a0}{(1)}^{\large n} \equiv 7 $   
