Induction prove $\begin{equation} 2^{n^2} \geq n! \end{equation}$ for a nonnegative integer

Question

Hi I submitted this as a graded assignment and received a poor grade. Could someone help me see what was wrong with my proof.

Let n be a nonnegative integer. Show that $\begin{equation}2^{n^2} \geq n!\end{equation}$

Proof

(i) Base Case

For n = 0 We have $\begin{equation}2^{0^2} \geq 0! \end{equation}$ Which Yields, $\begin{equation}1 \geq 1 \end{equation}$ Thus the base case holds.

(ii) Inductive Hypothesis:

Assume for some $\begin{equation} k\in\mathbb{Z}, k\geq 0 \text{ that }, 2^{k^2} \geq k!\end{equation}$ then look at $\begin{equation} k+1 \end{equation}$

\begin{align*} 2^{(k+1)^2} &= 2^{k^2 +2k+1}\\ &= 2^{k^2} \cdot 2^{2k} \cdot 2\\ &\geq k! \cdot 2^{2k} \cdot 2 \text{ via inductive hypothesis}\\ \end{align*}

We now take $\begin{equation} k!\cdot 2^{2k} \cdot 2 \end{equation}$ and relate it to $\begin{equation} (k + 1)! \end{equation}$

\begin{align*} k! \cdot 2^{2k} \cdot 2&\geq (k+1)!\\ k! \cdot 2^{2k} \cdot 2&\geq (k+1) \cdot k!\\ 2^{2k+1}&\geq (k+1)\\ \end{align*}

Thus the statement holds for $k+1$ Therefore by the generalized principle of mathematical induction,

$\begin{equation}2^{n^2} \geq n!\end{equation}$ for $\begin{equation} n\in\mathbb{Z}, n \geq 0 \end{equation}$

Answer 1

You need to prove that $2^{2k+1}\geq k+1$ and you did not do it.

For example: $$2^{2k+1}=(1+1)^{2k+1}\geq1+(2k+1)\cdot1>k+1$$