How to calculate last four digits of $2^{2017}$?

Since $10^4 = 2^4 \times 5^4$, my first reaction is one should use Chinese remainder theorem.

For $2^4$, it is clear $2^{2017} \equiv 0 \pmod {2^4}$.

For $5^4$, we use following theorem

For $a,b \in \mathbb{Z}_{+}$ such that $\gcd(a,b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(\cdot)$ is the Euler totient function.

For $a = 2$ and $b = 5^4$, we have $\varphi(5^4) = 4(5)^3 = 500$. This leads to $$2^{2017} = 2^{4(500)+17} \equiv 2^{17} \equiv 65536\times 2 \equiv 1072\pmod {5^4}$$

It turns out we are lucky. Since $1072 \equiv 0\pmod {2^4}$ already, we don't need CRT to conclude $2^{2017} \equiv 1072 \pmod {10^4}$. As a result, the last $4$ digits of $2^{2017}$ is $1072$.