Using Fermat's Little Theorem or Euler's Theorem to find the Multiplicative Inverse -- Need some help understanding the solutions here. 
The answers to multiplicative inverses modulo a prime can be found without using the extended Euclidean algorithm.
  a. $8^{-1}\bmod17=8^{17-2}\bmod17=8^{15}\bmod17=15\bmod17$
  b. $5^{-1}\bmod23=5^{23-2}\bmod23=5^{21}\bmod23=14\bmod23$
  c. $60^{-1}\bmod101=60^{101-2}\bmod101=60^{99}\bmod101=32\bmod101$
  d. $22^{-1}\bmod211=22^{211-2}\bmod211=22^{209}\bmod211=48\bmod211$ 

The above is using Fermat's little theorem to find the multiplicative inverse of some modular functions. However, there is a final step just before arriving at the answer that I do not understand how to solve, except to solve it by factoring. Factoring takes a very long time.
Basically, I don't see how the answers move from the third step to the fourth step aside from arriving at the answer by factoring. There has to be a better way using Fermat's Theorem or Euler's Theorem.

Bill's way seems great. Here's another approach (with the goal of finding easily reducible powers)
\begin{align}
8^{15}\pmod {17} &\equiv 2^{45}\\
&\equiv 2\cdot (2^{4})^{11}\\
&\equiv 2\cdot (-1)^{11} \tag{$16 \equiv -1$}\\
&\equiv 15 \tag{$15 \equiv -2$}
\end{align}

\begin{align}
5^{21}\pmod {23} &\equiv 5\cdot 5^{20}\\
&\equiv 5\cdot 25^{10}\\
&\equiv 5\cdot 2^{10} \tag{$25 \equiv 2$}\\
&\equiv 5\cdot 32^2\\
&\equiv 5\cdot 9^2 \tag{$32 \equiv 9$}\\
&\equiv 5\cdot 12 \tag{$81 \equiv 12$}\\
&\equiv 14 \tag{$60 \equiv 14$}
\end{align}

\begin{align}
60^{99}\pmod {101} &\equiv 10^{99}&\cdot 6^{99}\\
&\equiv 10\cdot 100^{49}&\cdot 6^4 \cdot (7776)^{19}\\
&\equiv -10&\cdot -6^4\\
&\equiv 12960\\
&\equiv 32
\end{align}

\begin{align}
22^{209}\pmod {211} &\equiv (2\cdot11)^{11\cdot19}\\
&\equiv ?\\
&\text{This is where the superiority of Bill's approach becomes obvious}
\end{align}
