Modular arithmetic problem (mod $22$) $$\large29^{2013^{2014}} - 3^{2013^{2014}}\pmod{22}$$
I am practicing for my exam and I can solve almost all problem, but this type of problem is very hard to me. In this case, I have to compute this by modulo $22$.

fermat's little theorem (and euler's theorem) is your friend.  
22 is coprime to 29 so $29^{\phi(22)} = 29^ {10} \equiv 1 \mod 22$.  So $29^{2013^{2014}} \equiv 29^k \mod 22$ where $2013^{2014} \equiv k \mod 10$.
As 10 and 2013 are co-prime $2013^{\phi (10)} = 2013^4 \equiv 1 \mod 10$ so$2013^{2014} \equiv 2013^2 \equiv 3^2 \equiv -1 \mod 10$.
So $29^{2013^{2014}} \equiv 29^{-1} \equiv 7^{-1} \mod 22$
As 22 is coprime to 3, by the exact same reasoning $3^{2013^{2014}} \equiv 3^{-1} \mod 22$.
So $29^{2013^{2014}} - 3^{2013^{2014}} \equiv 7^{-1} - 3^{-1} \equiv a \mod 22$.
Now $21a \equiv -a \equiv (7^{-1} - 3^{-1})7*3 \equiv 3 - 7 \equiv -4 \mod 22$.
So $29^{2013^{2014}} - 3^{2013^{2014}} \equiv  a \equiv 4 \mod 22$.
