Prove that $2^n < n!$, for all natural numbers $n > 3$
My proof:
Basis step: When $n = 4$, $2^4 = 16 < 4!=24$ which is true.
Inductive step: Assume that $2^k < k!$ for any $k > 3$. We must show that $2^{k+1} < (k+1)!$. Since $(k+1)! = (k+1)\cdot k!$, we have from the assumption that $$2 \cdot 2^k = 2^{k+1} < 2 \cdot k!.$$ Then we show that $2\cdot k! < (k+1)!$ when $k>3$ (so as to use transitive property of inequality to conclude $2^{k+1} <(k+1)!$). Note that $$ \begin{align} 2\cdot k! < (k+1)! &\iff2 \cdot k! < (k+1)\cdot k! \\ &\iff 2 < k + 1 \iff k>1 \end{align} $$ so since $k>3$ implies that $k>1$, it follows that $2\cdot k! < (k+1)!$. Hence from $2^{k+1} < 2 \cdot k!$ we get that $2^{k+1} <(k+1)!$. By the principle of mathematical induction, $2^n < n!$ is true for all natural numbers $n > 3$