Find $7^{100} \pmod{167}$ using Fermat's Little Theorem I understand how to use Fermat's Little theorem but only on exponents that are larger than the mod, therefore, I do not know how to do this question please help, thank you!

Since $\phi(167) = 166 = 2\cdot 83$, the possible orders of $7$ mod $167$ are $2$, $83$ and $166$.  Clearly, the order is not $2$.  If we compute the Legendre symbol 
$$\left(\frac{7}{167}\right) = - \left(\frac{167}{7}\right) = -\left(\frac{-1}{7}\right) = 1,$$
we see that $7$ is a quadratic residue mod $167$, and so the order can't be $166.$ (A primitive root can't be a quadratic residue.)
Thus, the order of $7$ mod $167$ is $83.$   So we have
$$7^{100} \equiv 7^{17} \pmod{167}.$$
Next, notice that $7^3 = 343$ is pretty close to $2\cdot 167 = 334$, so we
have $7^3 \equiv 9 \pmod{167}.$  Also, $2\cdot 9^2 = 162 \equiv -5 \pmod{167}$. Then we calculate:
$$7^{17} \equiv (7^3)^5\cdot 49 \equiv 9^2\cdot 9^2 \cdot 9 \cdot 49 \equiv (80\cdot 9^2 +1\cdot 9^2)\cdot 9 \cdot 49 \equiv (40(-5)+81)\cdot 9 \cdot 49 $$
$$\equiv (-119)\cdot 9 \cdot 49 \equiv -119\cdot 441 \equiv 48 \cdot (-60) \equiv -2880 \equiv 126 \pmod{167}.$$
