$4^{101} \mod 101$ without Fermat's little theorem Can anyone tell me how to find the remainder of $\frac{4^{101}}{101}$ without using Fermat's little theorem ? 
I tried doing $$4^{101} \equiv 2^{202} \equiv (5\cdot 101 + 7)^{22} \cdot 2^{4} \equiv 7^{22} \cdot 2^4 \mod 101$$
but after this I can't really get much further. 

This is how I would simplify it 
$$4^{101}=2^{202}\\2^{10}=1024\equiv 14\pmod{101}\\(2^{10})^2\equiv14^{2}\equiv-6\pmod{101}\\(2^{20})^5\equiv-6^5\equiv-216\cdot36\equiv-14\cdot36\equiv-504\equiv-1\pmod{101}\\(2^{100})^2\equiv 1\pmod{101}$$
