Better way to reduce $17^{136}\bmod 21$? What I have done:
Note $17\equiv -4$ mod 21, and $(-4)^2 \equiv 5$ mod 21. So $17^{136} \equiv (-4)^{136} \equiv 5^{68}$ mod 21. Also note $5^2 \equiv 4$ mod 21 and $4^3 \equiv 1$ mod 21, so $5^{68} \equiv 4^{34} \equiv (4^3)^{11}\cdot4 \equiv 4$ mod 21. I feel this is rather complicated, and there should be a better way. 

You could also reduce modulo each of the factors of $21$ and then use the Chinese Remainder Theorem to recover the result modulo $21$.
$$
  17^{136} \cong 2^{136} \cong (2^2)^{68} \cong 1^{68} \cong 1 \pmod{3}
$$
and
$$
  17^{136} \cong 3^{136} \cong 3^{6 \cdot 22 + 4} \cong (3^6)^{22} \cdot 3^4 \cong 1^{22} \cdot 3^4 \cong 3^4 \cong 9^2 \cong 2^2 \cong 4 \pmod{7} .
$$
Solving the system $x \cong 1 \pmod{3}$ and $x \cong 4 \pmod{7}$, we get $x \cong 4 \pmod{21}$. Therefore $17^{136} \cong 4 \pmod{21}$.
