Showing that for $n\geq 3$ the inequality $(n+1)^n<n^{(n+1)}$ holds

Solution 1:

Assume for contradiction that at some point we have $$ (n-1)^n>n^{n-1},\text{ but } (n+1)^n\geq n^{n+1} $$ and multiply the two inequalities.

Solution 2:

For the record, we can still do it by induction, it's just not as pleasant as other induction questions and the other proofs on this page.

Let's start with the LHS of the $n+1$ case. We're going to 'force in' the $n$ case, so we can use the induction hypothesis.

$$ \begin{align} (n+2)^{n+1} & = {(n+2)^{n+1}(n+1)^n \over (n+1)^n} \\ \\ & < {(n+2)^{n+1}n^{n+1} \over (n+1)^n} & \text{(induction hypothesis)} \end{align} $$

Now we work backwards from what we want to show to finish the induction step. We want:

$$\begin{align} {(n+2)^{n+1}n^{n+1} \over (n+1)^n} &< (n+1)^{n+2} \\\\ \iff (n+2)^{n+1}n^{n+1} &< (n+1)^{2n+2} \\\\ \iff [n(n+2)]^{n+1} &< [(n+1)^2]^{n+1} \\\\ \iff n(n+2) &< (n+1)^2 \end{align}$$

And this is true for all $n$ (simply expand it out). Hence we're done once we note it's true for $n=3$.

Alternatively if we know that $\lim_{n \to \infty} (1+1/n)^n = e < 3$, and that $(1+1/n)^n$ is increasing, we can observe that the inequality is true by dividing by $n^n$. But these facts require more work than a direct proof.

Solution 3:

For $n\geq 3, n\in\mathbb{N}$ we have $\binom{n}{2}=\frac{n(n-1)}{2}<n^2$ and so $(n+1)^n=\sum_{k=0}^n \binom{n}{k}n^{n-k}=1+\sum_{k=0}^{n-1} \binom{n}{k}n^{n-k}< 1+\sum_{k=0}^{n-1} n^kn^{n-k}=1+n\cdot n^n=1+n^{n+1}$ So we only need to show $n^{n+1}\neq (n+1)^n$. This is clear because one and only one of the numbers is odd.

Solution 4:

Hint Since $\log$ is an increasing function, taking the logarithm of both sides and rearranging gives that inequality is equivalent to $$\frac{\log n}{n} < \frac{\log (n + 1)}{n + 1}.$$ So, it suffices to show that the function $$x \mapsto \frac{\log x}{x}$$ is (strictly) decreasing on the interval $[3, \infty)$, which is a straightforward exercise using the First Derivative Test.

Solution 5:

Hint: rewrite as $n>\sqrt[n+1]{(n+1)^n}$ and try to use AM-GM inequality.

Further hint: $(n+1)^n=(n+1)^{n-1}\sqrt{n+1}^2$.

Further hint (per request): AM-GM gives us $$\sqrt[n+1]{(n+1)^n}=\sqrt{(n+1)\dots(n+1)\sqrt{n+1}\sqrt{n+1}}<\frac{(n+1)+\dots+(n+1)+\sqrt{n+1}+\sqrt{n+1}}{n+1}=\frac{(n-1)(n+1)+2\sqrt{n+1}}{n+1}$$