Can someone explain this asymptotic proof to me.I am stuck at the inductive step and get lost around this step $2 × n! < (n + 1) × n!$
$$2n = o(n!)$$
True Proof: In order to $2n = o(n!)$ be true, there should exist $n_0$ and a constant c such that $2^2n < c^2n$ starting $n_0$.
base case $n_0 = 4: (2^4 = 16) < (4! = 24)$.
Inductive step n + 1 case, assuming that $2n < n!$ for $n>4$ is true, show $2^n+1 < (n + 1)!$ $2 × 2^n < 2 × n!$ by assumption and $2 × n! < (n + 1) × n!$ since $n > 4$ $∴ 2^n = o(n!)$