<s>[INST] $k \in \mathbb Z^{+}$. Prove that for all $n ≥ 2k^2$, $n! ≥ k^{n}$ 
Proposition
$k \in \mathbb Z^{+}$. Then for all $n ≥ 2k^2$, $n! ≥ k^{n}$

My attempt:

Lemma 1. $k \in \mathbb Z^{+}$. For all $n \in \mathbb N$, $(k^2 + n)! ≥ k^{2n}$

Proof: By induction.
Base case: $n = 0$.
$(k^2 + 0)! ≥ 1  = k^{2\cdot0} = k^{2n}$
Induction step:
Suppose
$$(k^2 + n)! ≥ k^{2n}$$
Then
$$\begin{align}(k^2 + n + 1)! = (k^2 + n)! \cdot (k^2 + n + 1) & ≥ k^{2n} \cdot  (k^2 + n + 1) \\
& ≥ k^{2n} \cdot k \\
& = k^{2n+1}
\end{align}$$

Back to the proposition:
By induction.
Base case: $n = 2k^{2}$.
Applying lemma $1$, we conclude that $(2k^{2})! ≥ k^{2k^{2}}$
Induction step:
Suppose
$$n! ≥ k^{n}$$
Then
$\begin{align}(n+1)! = n! \cdot (n+1) & ≥ k^{n} \cdot (n+1)\\
& ≥ k^{n} \cdot k \space \space \space (\text{because } n > k)\\
& = k^{n+1}\\
\end{align}$
$\Box$
Is it correct?
 [/INST] You have made a mistake at the end of the proof of the lemma
\begin{eqnarray*}
(k^2 + n + 1)! = (k^2 + n)! \cdot (k^2 + n + 1) & ≥ k^{2n} \cdot  (k^2 + n + 1) \\
& ≥ k^{2n} \cdot \color{red}{k^2} \\
& = k^{2n+\color{red}{2}}.
\end{eqnarray*}
& in the base case of the proposition, you could be clearer by saying ... Using $n=k^2$ in the lemma gives $(2k^{2})!  \geq k^{2k^{2}} $.
</s>