Came up with this on my own and although it seems true (due to Desmos), I was interested to see a proof of it. I tried an inductive approach myself but unfortunately couldn't come up with anything concrete (just by assuming the statement, proving the base case and fiddling with it).
Prove that $$ (5^{k})! > 5^{k!} \ \forall \ k \in \mathbb{N} $$
For $k = 0$, we obtain $$ \left( 3^k \right)! = 1! = 1 \not> 5 = 5^1 = 5^{0!} = 5^{k!}. \tag{0} $$ For $k = 1$, we obtain $$ \left( 3^k \right)! = 3! = 3 \not> 5^1 = 5^{1!} = 5^{k!}. \tag{1} $$
For $k = 2$, we obtain $$ \left( 3^k \right)! = 9! > 25 = 5^2 = 5^{2!} = 5^{k!}. \tag{2} $$
Suppose that $k \in \mathbb{N}$ such that $k \geq 2$ and also $$ \left( 3^k \right)! > 5^{k!}. \tag{3} $$
Then we find that $$ \begin{align} \left( 3^{k+1} \right)! &= \left( 3 \cdot 3^k \right)! \\ &= \left( 3 \cdot 3^k \right)\left( 3 \cdot 3^k -1 \right) \ldots \left( 3^k +1 \right) \left( \cdot 3^k \right)! \\ &> \left( 3 \cdot 3^k \right)\left( 3 \cdot 3^k -1 \right) \ldots \left( 3^k +1 \right) 5^{k!} \\ &= \left( 3^k + 2 \cdot 3^k \right) \left( 3^k + 2 \cdot 3^k -1 \right) \ldots \left( 3^k + 1 \right) 5^{k!} \\ &> \left( 3^k + 1 \right)^{2 \cdot 3^k} 5^{k!} \\ &> 5^{2 \cdot 3^k } 5^{k!} \\ &> 5^{k+1} 5^{k!} \tag{4} \\ &= 5^{(k+1)k!} \\ &= 5^{(k+1)!}. \end{align} $$
In (4) above here we have used the result that $$ 2 \cdot 3^k > k+1 $$ for all $k \in \mathbb{N}$. This result should not be too hard to prove using induction.
Hope this helps.