Prove by induction $n!<4n^2+3$ How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?
Here´s my try:
The equation is valid for every $n\leq4$.
n=5: n!=5!=120>103
n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$
Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.
Any ideas how i could continue or whats wrong?

Assuming you want to prove that for each $n\geq 5$ one has $n! > 4n^2+3$ just note that it easily holds for $n=5$. Now assume that it holds up to $n \geq n-1 \geq \ldots 5.$ Then
$$(n+1)! = (n+1)n! > (n+1)(4n^2+3).$$
It is only left to prove that $(n+1)(4n^2+3) > 4(n+1)^2+3$ for each $n\geq 5.$ In fact,
$$(n+1)(4n^2+3)-4(n+1)^2-3 = 4n^3-5n-4.$$
So let $p(n) = 4n^3-5n-4.$ Note that $p(5) > 0.$ Moreover, $p'(n) = 12n-5 > 0,~\forall n\geq 5.$
