Three years, back and forth

Consider $f(2015)$ , which we will write as $2015^{2016^a}$ . To find the last four digits of this value, we need to find residues $x$ and $y$ such that $x \equiv 2015^{2016^a} \pmod {16} \qquad \text{and} \qquad y \equiv 2015^{2016^a} \pmod {625}.$ By the Chinese remainder theorem, these two residues will tell us the value of $2015^{2016^a}$ modulo $16 \cdot 625 = 10000$ . Both residues are easy to find. Notice that $2015^2 \equiv 1 \pmod {16}$ . Since $2016^a$ is clearly even, we have $2015^{2016^a} \equiv 2015^{2k} \equiv (2015^2)^k \equiv 1^k \equiv 1 \pmod{16}.$ Also, notice that $5$ is a factor of $2015$ . Since $2016^a$ is very large, $5^4 = 625$ is a factor of $2015^{2016^a}$ and thus $2015^{2016^a} \equiv 0 \pmod{625}.$ By the Chinese remainder theorem, the previous two congruences imply $2015^{2016^a} \equiv 625 \pmod {10000},$ so the last four digits of $2015^{2016^a}$ are $0625$ .

Now consider $g(2021)$ , which we will write as $2021^{2020^b}$ . Notice that 2021 and 10000 are relatively prime and $\phi(10000) = 4000$ . If $2020^b \equiv r \pmod{4000}$ , then Euler's theorem implies $2021^{2020^b} \equiv 2021^r \pmod{10000}.$ The prime factorization of 4000 is $2^5 \cdot 5^3$ , and $2$ and $5$ are both factors of 2020. Since $b$ is very large, $2^5$ and $5^3$ are both factors of $2020^b$ and thus $2020^b \equiv 0 \pmod{4000}.$ By Euler, $2021^{2020^b} \equiv 2021^0 \equiv 1 \pmod{10000},$ so the last four digits of $2021^{2020^b }$ are $0001$ .

Therefore, the desired value is $625 + 1 = \boxed{626}$ .