Prove by induction that $2^n\le (n+1)!$ for all $n\in\Bbb N$.

Question

Prove by induction that $2^n\le (n+1)!$ for all $n\in\Bbb N$.

What I've done so far is prove for $n=1$:

$$2^1 \le (1+1)!,$$ $$2 \le 2,$$ which is correct

Then I tried to prove for $n+1$, in other words, I want to get here: $$2^{n+1}\le(n+2)!.$$

So, I multiplied everything with $2$: $$2\cdot 2^n \le2\cdot (n+1)!,$$ $$2^{n+1} \le2(n+1)n!.$$

So I already have what I wanted in the left part of the inequality, but I'm stuck for the right part.

Can someone help me? Thanks!

Answer 1

hint if n is a positive integer then $(n+2)!\geq (n+1)!$so that $(n+2)(n+1)!\geq2(n+1)!$

Btw that 2 comes from an easy proof by developping the parenthesis. Now use

($a<b<c\implies a<c$)

In your second inequality

Answer 2

Hint: $$2\,(n+1)!<(n+2)\,(n+1)!=(n+2)!$$

Answer 3

Since a hint has already been provided, I am supplying a combinatorial approach. There are $(n+1)!$ permutations on the set $\{0,1,2,\ldots,n\}$. Amongst these, consider permutations $\sigma$ such that, for any $k=0,1,2,\ldots,n$, there exists $j_k\in\{0,1,2,\ldots,n-k\}$ such that $\sigma$ maps $\{0,1,2,\ldots,k\}$ to $\left\{j_k,j_k+1,\ldots,j_k+k\right\}$. Show that there are $2^n$ such permutations.

An Example: Take $n=2$, and represent permutations as juxtapositions of labels $0$, $1$, and $2$. Then, amongst all six permutations $012$, $021$, $102$, $120$, $201$, and $210$, only four of them ($012$, $102$, $201$, and $210$) are the special permutations we consider.