Suppose $5^{4a} + 5^{3a} + 5^{2a} + 5^a + 1$ is composite for some $a \geq 1$. Then for every prime $p$ dividing that expression we have:
$$5^{4a} + 5^{3a} + 5^{2a} + 5^a + 1 \equiv 0 \pmod{p}$$
That is:
$$\frac{5^{5a} - 1}{5^a - 1} \equiv 0 \pmod{p}$$
Which implies that:
$$5^{5a} \equiv 1 \pmod{p} ~ ~ ~ \text{and} ~ ~ ~ 5^a \not \equiv 1 \pmod{p}$$
Observe that these equations imply $5^a$ has order 5 modulo $p$, since 5 is prime. Now let $k$ be an integer not multiple of 5, then we have:
$$\left ( 5^{5a} \right )^k \equiv 5^{5ak} \equiv 1 \pmod{p}$$
And since $k$ is not a multiple of 5 and $5^a$ has order 5 modulo $p$, it must follow that:
$$\left ( 5^{a} \right )^k \equiv 5^{ak} \not \equiv 1 \pmod{p}$$
So we conclude that:
$$\frac{5^{5ak} - 1}{5^{ak} - 1} \equiv 0 \pmod{p}$$
Such that:
$$5^{4ak} + 5^{3ak} + 5^{2ak} + 5^{ak} + 1 \equiv 0 \pmod{p}$$
And so we conclude that $f(ak)$ is a multiple of $f(a)$ and so is also composite. With the above theorem it suffices to show that the expression is composite for powers of 5. In particular, it is composite at $a = 1 = 5^0$, hence is composite at all $a$ not multiple of 5, and it can be verified that it is also composite at $a = 5$, hence it is composite at all $a$ not multiple of 25. We can keep going and check that the expression is composite for $a = 25$ as well, showing that the expression is composite for all $a$ not multiple of 125, but I don't know how to prove that it must be composite for all powers of 5, which is required to show it is composite for all integer $a$.