Induction proof that $n! > n^3$ for $n \ge 6$, and $\frac{(2n)!}{n! 2^n}$ is an integer for $n \ge 1$ Prove by induction that
(a) $n! > n^3$ for every $n \ge 6$.
(b) prove $\frac{(2n)!}{n!2^n}$ is an integer for every $n\geq 1$
I'm quite terrible with induction so any help would be appreciated. 

(a) $n! > n^3$ for every $n \geq 6$
For the induction base, we simple have to show that $6! \geq 6^3$. Hence, $6 * 5 * 4 * (3 * 2) * 1 \geq 6*6*6$. Hence $6 * 6 * (5 * 4) \geq 6^3$. This is evidently true, as $5*4 = 20 > 6$.
For the induction step, we have to show that $(n+1)! \geq (n+1)^3$ for $n\geq 6$. Now suppose $n! > n^3$ is true for all $n \geq 6$. Then $(n+1)! = (n+1)n! \geq (n+1)n^3$.
Hence, we are done if $(n+1)n^3 \geq (n+1)^3$ for $n\geq 6$. Thus, we need to show that $n^4 + n^3 \geq n^3 + 3n^2 + 3n + 1$ for $n\geq 6$.
Rearranging yields $n^4 - 3n^2 - 3n - 1 \geq 0$ for $n\geq 6$. The second derivative of this is $12n^2 - 6$, which is strictly convex for $n\geq 6$. Hence, we are through if both $f(n) := n^4 - 3n^2 - 3n - 1 \geq 0$ and $f'(n) = 4n^3 -6n - 3 \geq 0$ for $n=6$. This is trivial to check.
(b) $\frac{(2n)!}{n! 2^n}$ is an integer for every $n\geq 1$.
The induction base is trivial: simply insert $n=1$.
For the induction step, we need to show that $\frac{(2(n+1))!}{(n+1)^2 2^{n+1}} = \frac{(2n+2)!}{(n+1)n! * 2 * 2^n} = \frac{(2n+2)(2n+1)}{2(n+1)} * \frac{(2n)!}{n! 2^n}$ is an integer for all $n\geq 1$.
Now, by the induction hypothesis, the right multiplicand must be an integer. Hence, it suffices to show that the left multiplicand must be an integer as well. Since $(2n+2) = 2(n+1)$, this is obviously the case.
