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?

Your work is mostly fine, but you can perfect your redacting.
What you want to show is that the set $A=\{n\in\mathbb N\mid n!<4n^2+3\}=\{0,1,2,3,4\}$
For that you need to verify the inequality for the a few numbers which you certainly did, but it needs to appear on your solution.
$\begin{cases}
n=0&0!=1<4\times 0^2+3=3&\text{ok}\\
n=1&1!=1<4\times 1^2+3=7&\text{ok}\\
n=2&2!=2<4\times 2^2+3=19&\text{ok}\\
n=3&3!=6<4\times 3^2+3=39&\text{ok}\\
n=4&4!=24<4\times 4^2+3=67&\text{ok}\\
n=5&5!=120\not<4\times 5^2+3=103&\text{not verified}\\
\end{cases}$
You are correctly suspecting that the inequality is not verified either for $n\ge 5$, so you should explicit the induction proposition :
$$P(n) : n! \ge 4n^2+3\qquad\forall n\ge 5$$
Base case $P(5)$ is verified.
So let assume $P(n)$ we get $(n+1)!=(n+1)n!\ge (n+1)(4n^2+3)$
And this is where you got stuck.
You need to make appear $4(n+1)^2+3$, instead I propose to subtract both quantities and show the difference is positive (this is easier to handle):
$\begin{align}(n+1)!-\bigg((4(n+1)^2+3\bigg)
&\ge (n+1)(4n^2+3)-\bigg(4(n+1)^2+3\bigg)
\\&\ge 4n^3+3n+4n^2+3-(4n^2+8n+7)
\\&\ge 4n^3-5n-4
\\&\ge 4n^3-5n-4n\qquad\text{ for } n\ge 1\implies 4\le 4n
\\&\ge n(4n^2-9)
\\&\ge 0 \qquad\qquad\qquad\qquad\text{ for } n^2\ge \frac 94\iff n\ge \frac 32
\end{align}$
Since $n\ge 5$ both conditions ($n\ge 1$ and $n\ge \frac 32$) are verified therefore you have proved your induction step.
