Find the last two digits of $2014^{2014}$.
(This number has 6655 digits.)
The last two digits are the least non-negative residue modulo 100.
We have $2014 \equiv 14 \pmod{100}$, and $14^2 = 196 \equiv -4 \pmod{100}$.
By Proposition (3.8) we have
$$ \begin{align} 2014^{2014} & \equiv (14)^{2014} \pmod{100} \\ \\ & \equiv (14)^{2 \times 19 \times 53} \pmod{100} \\ \\ & \equiv (14^2)^{19 \times 53} \pmod{100} \\ \\ & \equiv (-4)^{19 \times 53} \pmod{100} \end{align} $$
Now, $19 \times 53$ can't be decomposed further, so we'll resort to the division algorithm.
$$ 19 \times 53 = 1007 = (10 \times 100) + 7$$
And so,
$$ \begin{align} 2014^{2014} & \equiv (-4)^{(10 \times 100) + 7} \pmod{100} \\ \\ & \equiv (-4)^{10 \times 100} \times (-4)^7 \pmod{100} \\ \\ & \equiv ((-4)^{10})^{100} \times (-4)^7 \pmod{100} \end{align} $$
It is now feasible to calculate
$$ (-4)^{10} \equiv -1048576 \equiv 24 \pmod {100} $$
$$ (-4)^7 \equiv -16384 \equiv 16 \pmod {100} $$
And so,
$$ \begin{align} 2014^{2014} & \equiv (24)^{100} \times 16 \pmod{100} \end{align} $$
We can repeat this process using $24^{100} = (24^{10})^{10} = (6^{10} \times 4^{10})^{10} \equiv (76 \times 76)^{10} \equiv 76^{10} \pmod{100}$
$$ \begin{align} 2014^{2014} & \equiv (24^{10})^{10} \times 16 \pmod{100} \\ \\ & \equiv (76)^{10} \times 16 \pmod{100} \end{align} $$
Similarly $76^{10} = 4^{10} \times 19^{10} \equiv 76 \times 1 \pmod {100}$, and so
$$ \begin{align} 2014^{2014} & \equiv76 \times 1 \times 16 \pmod{100} \\ \\ & \equiv 1216 \pmod{100} \\ \\ & \equiv 16 \pmod{100} \end{align} $$
And so the last two digits of $2014^{2014}$ are 16.