Prove that $(n! +n) \leq (n+1)!$

Question

How can I prove that $(n! +n) \leq (n+1)!$ given that $n \geq 0$?

Answer 1

If $n \ge 1$, this is equivalent to $$ 1 \le n(n-1)! $$ which is clear.

Answer 2

$\frac{1}{n+1} + \frac{n}{(n+1)!} \leq \frac{1}{n+1} + \frac{n}{(n+1)} = 1$

Therefore since $\frac{1}{n+1} + \frac{n}{(n+1)!}= \frac{n!}{(n+1)!} + \frac{n}{(n+1)!} $ your inequality follows immediately.

Answer 3

Also if $n=0$ it is obvious ($1\le 1$),

for $n\ge 1$: $$n!+n\le n!+n!=n!2\le (n+1)!$$

Answer 4

$n!+n\le n!+n\cdot n!=n!(1+n)=(n+1)!\;$

for all $\;n\in\mathbb{N}\cup\left\{0\right\}.$