Prove the statement by the Principle of Mathematical Induction:
$2n < (n+2)!$ for all natural numbers $n$.

(N/A) Let $P(n): 2n < (n+2)!$ for all $n \in \mathbb{N}$.
Step $1$: For $n=1$,$P(1): 2(1) < (1+2)! \implies 2 < 3! \implies 2 < 6$,which is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$2k < (k+2)!$.
Step $3$: We need to prove $P(k+1): 2(k+1) < (k+3)!$.
Starting from the assumption $2k < (k+2)!$,we add $2$ to both sides:
$2k + 2 < (k+2)! + 2$
$2(k+1) < (k+2)! + 2$
Since $(k+2)! + 2 < (k+2)! \times (k+3)$ for all $k \ge 1$ (because $(k+2)! \times (k+3) - (k+2)! = (k+2)!(k+3-1) = (k+2)!(k+2) \ge 2! \times 2 = 4 > 2$),
we have $2(k+1) < (k+2)! + 2 < (k+3)!$.
Thus,$2(k+1) < (k+3)!$,which means $P(k+1)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.