Proof by Induction - Am I doing this correctly? 
For all n that is a natural number where n ≥ 4, n! > n^2

We wish to prove by induction.
Let us call (n! > n^2) p(n).
Base Case: We will test out p(4).
4! = 24
$4^2 = 12.$ As 24 > 12, p(4) is proven.
Induction Hypothesis and Conclusion:
Assume $n! > n^2$ for all $n ≥ 4$
We wish to prove p (k+1). That is, $((n+1)! > (n+1)^2)$
$(n+1)! ≥ n!(n+1) > n^2(n+1)$ using the Induction Hypothesis
$n^2(n+1) ≥ (n+1)^2$ occurs when $n^2 ≥ (n+1)$
$ n^2 ≥ (n+1) $
$ n ≥ 1 + 1/n $
As n ≥ 4, this is true. There for, we have shown p(n+1) and proved the original statement.
 A: There is a small problem: $4^2=16$ (not $12$).
And there is a big problem, which is when you write “Assume $n! > n^2$ for all $n ≥ 4$”. If you assume that, then you will be assuming the very thing that you wish to prove. What you should do is to take some $n\in\mathbb N$, to assume that the inequality $n! > n^2$ holds for that $n$ and to prove that then $(n+1)!>(n+1)^2$.
A: Seems good to me but for some minor mistakes:
Assume $n! \gt n^2$ for some $n$, not for all. With p(4), you have shown the statement only for one value.
You want to write $(n + 1)$ instead of ($n = 1$) in the second to last line and you can omit the last line in my opinion, since this inequality is clear.
A: Your logic and plan is absolutely correct but your execution has some bugs.
First off there is the silly bit that $4^2 = 16\ne 122$.... Oh, well.  That's trivial.
Then there is the the "Assume $n! \ge n^2$ for all $n\ge 4$".  That's actually what you want to prove.  In the induction step you want to assume $n! \ge n^2$ for one value of $n \ge 4$.
I see a lot of proofs that use "Assume this is true for $n =k$ for some value of $k \ge 4$".  There is an advantage to this in that it doesn't confuse the $n$ representing any value as it does in the statement and $n$ being a specific value as it does in the induction step.
$n^2 \ge (n=1)$.   I actually was not able to read this and honestly did not know what you meant. I finally realised it was a typo and you mean $n^2 \ge (n+1)$
The lines:
1 $n^2(n+1) \ge (n+1)^2$ whenever $n^2 \ge (n+1)$
2 $n^2 \ge (n+1)$
3 $n \ge 1 + \frac 1n$
It isn't clear what the logic behind this is supposed to say.  It looks like you are saying $n^2 \ge (n+1)$ as a fact and we don't know why you can say that.  And it looks like your conclusion is $n \ge 1 + \frac 1n$. And we don't know why we care about that.
You should somehow say "$n^2 \ge (n+1) \iff n \ge 1 + \frac 1n$; and that is our case because $n \ge 4$".  You could also add a line about why $n \ge 4$ means $n > 1 + \frac 1n$, but that's not a deal breaker.  ($1 + \frac 1n < 2$ so $n> 1+ \frac 1n$ will be true for all $n \ge 2$.)
A: We can make it slightly simpler by noting that
$$n!>n^2\iff (n-1)!>n.$$
Then,
$$n!=n(n-1)!>n^2>n+1$$
because $n^2-n=n(n-1)>1$ for $n>1$.
