Find the last digit of $7^{7^{7^7}}$ I was trying to solve this equation but I couldn't find a way to prove that this number ends with a 3 (my teacher has given me the answer).
Can someone explain me why?
$7^{7^{7^{7}}}$

Note that $7^{2}$ ends with a $9$, and thus $7^{3}$ ends with a $3$ --- the last digit of $7\cdot 9$. Playing more with this we discover that the last digits of potencies of seven are $7,9,3,1,7,9,3,1,\dots$. So $7^{7}$ ends with a $3$. What is the last digit of the seventh power of a number ending in $3$? i.e. $7^{7^{7}}$. What about this number's seventh power? 
