Inequality with heterogeneous functions I need to prove that $$2^n - n! \leq 0.$$ How do I calculate (without hit and trial) the value of $n$ ?

Try a proof by induction starting with $n=4$. When $n=4$, we have $2^4=16 < 4!=24$ so the inequality holds. Next, suppose that $2^n \leq n!$. Then multiplying both sides of this inequality by two, we find $2^{n+1}\leq 2\cdot n! \leq (n+1) n! = (n+1)!$ which completes the proof.