Find the last digit of $7^{2014}$.
We remind ourselves of Euler's Theorm (5.14), which states that for $n>1$ and $\gcd(n.a)=1$, then
$$ a^{\phi(n)} \equiv 1 \pmod n $$
Since $\gcd(10,7)=1$ we have
$$ 7^{\phi(10)} \equiv 1 \pmod {10} $$
That is
$$ 7^{4} \equiv 1 \pmod {10} $$
Using $2014 = 4(503)+2$, we have
$$ \begin{align} 7^{2014} & \equiv (7^4)^{503} \times 7^2 \pmod {10} \\ \\ & \equiv 1 \times 49 \pmod {10} \\ \\ & \equiv 9 \pmod {10} \end{align} $$
And so the last digit of $7^{2014}$ is 9.