1. Let $p$ be a prime number greater than 5. Assume for the sake of contradiction that $p - 4$ is the fourth power of a prime number. Let $q$ be a prime number such that $p - 4 = q^4$.

2. Then we have:
$$p = q^4 + 4$$

3. We can rewrite $q^4 + 4$ using the Sophie Germain identity:
$$q^4 + 4 = (q^2 + 2)^2 - (2q)^2 = (q^2 - 2q + 2)(q^2 + 2q + 2)$$

4. Since $q$ is a prime number and $q > 1$, both factors $q^2 - 2q + 2$ and $q^2 + 2q + 2$ are greater than 1.

5. For $p$ to be a prime number, one of the factors must be equal to 1. However, we will show that neither $q^2 - 2q + 2$ nor $q^2 + 2q + 2$ can be equal to 1.

6. Consider the factor $q^2 - 2q + 2$:
$$q^2 - 2q + 2 = 1 \implies q^2 - 2q + 1 = 0 \implies (q-1)^2 = 0 \implies q = 1$$
But $q = 1$ is not a prime number, so $q^2 - 2q + 2 \neq 1$.

7. Now consider the factor $q^2 + 2q + 2$:
$$q^2 + 2q + 2 = 1 \implies q^2 + 2q + 1 = 0 \implies (q+1)^2 = 0 \implies q = -1$$
But $q = -1$ is not a prime number, so $q^2 + 2q + 2 \neq 1$.

8. Since neither $q^2 - 2q + 2$ nor $q^2 + 2q + 2$ can be equal to 1, both factors are greater than 1. This implies that $p$ is a product of two integers greater than 1, which contradicts the assumption that $p$ is a prime number.

Therefore, $p - 4$ cannot be the fourth power of a prime number.

$\blacksquare$