Proof (cases & induction): Find the set of positive integers such that $n! \geq n^3$ I need to find the set of positive integers such that $n! \geq n^3$, and then prove my answer is true using cases and induction on $n$.
There is a lemma that I will need to prove and use for this proof. 
The lemma is : 
$n^2+2n+1\leq n^3$ when $n\geq$ ??

Here is my outline of the proof:
Claim: The set of all positive integers that satisfy $n! \geq n^3$ is $\{n\in \mathbb{Z} | n=1$ or $n\geq 6\}$
Let us prove the following Lemma : $n^2+2n+1\leq n^3$ when $n\geq 3$
Let $n \in \mathbb{Z}^+$
$n^2+2n+1 \leq n^2+2n+n$ when $n\geq 1$
$n^2+2n+1 \leq n^2+n^2+n$ when $n\geq 2$
$n^2+2n+1 \leq n^2+n^2+n^2$ when $n \geq 3$
$n^2+2n+1 \leq 3n^2 \leq n^3$ when $n \geq 3$
Thus, we have $n^2+2n+1\leq n^3$ when $n\geq 3$
This lemma is important because when we use induction on $n$, we want to eventually show that 
$n!\geq n^3$ implies $(n+1)!\geq(n+1)^3$
In order to show this, we need to multiply both sides of $n!\geq n^3$ by $(n+1)$
Our result would be $(n+1)n!\geq (n+1)n^3$
Now we need to show that $(n+1)n^3\geq (n+1)(n+1)^2$ so that we can get to our goal.
Using our lemma, we have proved that $(n+1)^2\leq n^3$ when $n\geq 3$, so we know that we can use induction on our original claim so long as $n\geq 3$.
So using our lemma, we can assume that if $(n+1)n!\geq (n+1)n^3$ is true, then $(n+1)!\geq(n+1)^3$ is also true, so long as $n\geq 3$
Now we can use cases and induction to prove our claim.
Case 1: $n=1$, $1!\geq 1^3$
Case 2: $n=2$, $2!\ngeq 2^3$
Case 3: $n=3$, $3!\ngeq 3^3$
Case 4: $n=4$, $4!\ngeq 4^3$
Case 5: $n=5$, $5!\ngeq 5^3$
Case 6: $n=6$, $6!\geq 6^3$
Let us use induction on $n$ to prove that $n! \geq n^3$ for all positive integers greater than or equal to $6$.
Base case: $n=6$. $6!\geq 6^3$, so we have proved the base case.
Inductive step: Suppose we have $(\star)$ $n!\geq n^3$. We want to show that $(n+1)!\geq(n+1)^3$. Let us multiply both sides of $(\star)$ by $(n+1)$.
We have $(n+1)n!\geq (n+1)n^3$ This can be rewritten as $(n+1)!\geq (n+1)n^3$
Using our lemma, we can assume that if $(n+1)!\geq (n+1)n^3$ is true, then $(n+1)!\geq(n+1)^3$ is also true, so long as $n\geq 3$.
This completes the induction, and we have proved the claim.

My questions are: 
1) Have I completely proved the lemma?
2) Are there any errors in the proof?
3) How can I improve this proof?

Let $f(x) = x^3 - (x + 1)^2$ 
$f'(x) = 3x^2 - 2x + 2$.  $f'(x) = 0$ when $x = \frac{2 \pm \sqrt(4 - 24}{6}$ so $f'(x)$ never equals $0$ for any real x.  So $f(x)$ has no extrema and thus only one real root.
$ f(2) = -1; f(3) = 18$ so $f(x)$ has only one  real root is between 2 and 3.  So $f(n) > 0$ iff $n^3 > (n+1)^2$ iff $n \ge 3$.
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D'oh!  The obvious easiest way:
Let $ m = n+1;$  $\frac{n^3}{n^2 + 2n + 1} = \frac{(m - 1)^3}{m^2}$
$= \frac{m^3 - 3m^2 + 3m -1}{m^2} = m - 3 + (\frac 3 m - \frac 1 {m^2})> m -3$
so  $\frac{n^3}{n^2 + 2n + 1} > n -2$
so $\frac{n^3}{n^2 + 2n + 1} > 1 \iff n^3 > n^2 + 2n + 1$ when $n \ge 3$.
