Find the last three digits of $27^{1 000 000}$.
Since $\gcd(27,1000)=1$, Euler's Theorem tells us
$$ 27^{\phi(1000)} \equiv 1 \pmod {1000} $$
Using $\phi(1000) = 400$ gives us
$$ 27^{400} \equiv 1 \pmod {1000} $$
Noting that $1000000 = 400 \times 2500$, we have
$$ 27^{1000000} \equiv (27^{400})^{2500} \equiv 1^{2500} \equiv 1 \pmod {1000} $$
And so the last three digits of $27^{1000000}$ are $001$.