Prove $2^n > n$ and if $2^{n+1}>n!$.
I am currently studying noetherian rings and found (quite embarrassingly) that I have forgotten basic proofs as such. I want to know if my proofs are airtight.
For $2^n >n$, I use induction. Base case is obvious, and I assume for $n$ and prove for $n+1$. We prove $2^{n+1} = 2 \cdot 2^{n}$ and since $2^n > n $ then $2 \cdot 2^n > 2n = n+n > n+1$.
I feel there was an unnecessary step. For the second proof, base case is also obvious. I assume for $n-1$ and prove for $n$. Thus, $2^{n-1}>(n-1)!$ which implies $2^n > 2(n-1)!$ but this is wrong because $2(n-1)! \not\ge n!$. I then show that $x!$ eventually overtakes $2^{x+1}$ at almost $x = 4.33...$ but I got curious and would like to find an exact algebraic formula where $2^{x+1}=x!$. I know there are multiple points where they intersect, especially behind $x=0$, but how would I find these solutions?
Thanks in advance.