Proof that $ n^{n} \leq {(n+1)}^{n} $

Question

I know this seems trivial, but how could I proof this? Should I use Induction? Where $n$ is an integer.

Answer 1

Using the binomial theorem,

$$ (n+1)^n=\sum_{i=0}^n\binom{n}{i}n^i. $$

One of the terms is $n^n$ and the rest are positive, so $(n+1)^n$ is $n^n+$something positive, and the inequality follows.

Answer 2

$$ (n+1)^n = [n(1+1/n)]^n = n^n(1+1/n)^n \geq n^n\cdot1^n = n^n $$

whenever $n \in \mathbb{N}$ is greater than or equal to $1$.

Answer 3

From Andre's hint in the comments:

$$n \leq n + 1$$

In fact, strict inequality holds:

$$n < n + 1$$

Assuming $n \in \mathbb{N}$ (since I see that you've mentioned induction), we can raise both sides of the last inequality to the $n$th power:

$$n^n < (n + 1)^n.$$

This implies that

$$n^n \leq (n + 1)^n,$$

since $\leq$ is to be interpreted as less than OR equal to.

QED

Answer 4

Induction is a good and simple way to prove $$a \lt b \implies a^n<b^n$$ for $ a,b,n \in \mathbb{N}$ because $$a^{n+1}=a^n\cdot a<b^n \cdot b=b^{n+1}$$ From this your statement follows.