I assume you are talking about $n\geq3$, otherwise your proposition is false. Your proof could be done by induction. The base case, $n=3$, is trivially shown. For the inductive step, let us consider the functions: $$f(k)=(1+\frac 1k)^k$$ and $$g(k)=k+1$$ Try to show that for all $k\geq 3$, $g(k)>f(k)$. For this, you can use calculus. Notice that both $f$ and $g$ are monotonically increasing and both $g(3)>f(3)$ and $\lim_{k\to \infty}g(k)>\lim_{k\to \infty}f(k)$. Once you've proven this, if we take a statement $S_k$ as: $"(k!)^2>k^k"$. Then, $$k+1>(1+\frac 1k)^k$$ So, $$(k+1)k^k>(k+1)^k$$ So, $$k^k>(k+1)^{k-1}...(1)$$ Which means, $$(k+1)^2 k^k>(k+1)^{k+1}$$ Now, multiplying $(k+1)^2$ on both sides of $S_k$, we get: $$((k+1)!)^2>(k+1)^2 k^k...(2)$$ Combining $(1)$ and $(2)$, you get that $S_{k+1}$ is also true.

QED.

I assume you are talking about $n\geq3$, otherwise your proposition is false. Your proof could be done by induction. The base case, $n=3$, is trivially shown. For the inductive step, let us consider the functions: $$f(k)=\left(1+\frac 1k\right)^k$$ and $$g(k)=k+1$$ Try to show that for all $k\geq 3$, $g(k)>f(k)$. For this, you can use calculus. Notice that both $f$ and $g$ are monotonically increasing and both $g(3)>f(3)$ and $\lim_{k\to \infty}g(k)>\lim_{k\to \infty}f(k)$. Once you've proven this, if we take a statement $S_k$ as: "$(k!)^2>k^k$". Then, $$k+1>\left(1+\frac 1k\right)^k$$ So, $$(k+1)k^k>(k+1)^k$$ So, $$k^k>(k+1)^{k-1}...(1)$$ Which means, $$(k+1)^2 k^k>(k+1)^{k+1}$$ Now, multiplying $(k+1)^2$ on both sides of $S_k$, we get: $$((k+1)!)^2>(k+1)^2 k^k...(2)$$ Combining $(1)$ and $(2)$, you get that $S_{k+1}$ is also true. QED.