Are the statements: $n!=O(2^n)$ and $2^n n^2=O(3^n)$ true or false? I am trying to do this exercise: 
Are the statements: a)$\space n!=O(2^n)$ and b)$\space2^nn^2=O(3^n)$ true or false? 
I am trying to show that a) is false and still don't know what to say about b). This is what I've tried to do up to now:
a) I could prove by induction the inequality $n!>2^n$ for all $n \geq 4$. I am trying to deduce from here that there are no natural numbers $c$ and $n_0$ such that for all $n \geq n_0$, $n! \leq c2^n$ but I couldn't do it.
b) I know that $2^n \leq 3^n$ and that $n^2 \leq 3^n$. Now, from these two inequalities, I can deduce that $2^nn^2 \leq 3^{2n}=9^n$, but I can't conclude from here that there is a constant $c$ such that $2^nn^2 \leq c3^n$ for sufficiently large $n$.
I would also like to understand intuitively for $b)$ why is it that the inequality is true, in case it is actually true, or why it is false, in case the inequality turns out to be false. 
Thanks in advance.

For the first part,
you can go by doing something like this. 
You can get power of prime $x$ in $n!$ by $F(x,n)=\sum \lfloor \frac {n}{x^i} \rfloor$, where $F(x,n)$ is power of $x$ in $n!$. This is called Legendre's Formula.
Now express $n!$ as product of prime powers , $n=\prod i^{F(i,n)}$ where $i$ is a prime. This representation is called De Polignac's formula.
Even if we assume that we can eliminate $2^n$ from $n!$, $\frac {n!}{2^n}$ will still remain a function of $n$ as there are other prime powers which are function of $n$. So you can easily  prove that there are no such $c$ and $n_0$.
For second part, express $3^n$ = $(1 + 2)^n$.
So we can say using Binomial expansion that,
$3^n= \frac{n^2\cdot2^n}{8}$ + some positive number (part of third term in binomial expansion + some positive number)
Clearly, without that other positive number,
$n^2\cdot2^n < 8\cdot3^n.$ So $n^2\cdot2^n$  is O($3^n$).
