Show that eventually $(n+2)^n < (n+1)^{n+1}$
I can see that this is obvious by evaluation at n>2, but I am having a hard time separating to get the induction step within the parenthesis. I am sorry if this is too easy a problem.
Or, maybe there's a way without induction?
Divide both sides by $(n+1)^n$. Then you have $(1 + \frac{1}{n+1})^n < n+1$. But since the left side is less than $e$, the inequality holds for all $n > e - 1$.
You can show that $(1 + \frac{1}{n+1})^n < e$ by showing that it increases monotonically for positive numbers (take the derivative).
Alternative approach: your inequality is equivalent to $\frac{1}{n+2}<[(1+n)/(n+2)]^{n+1}$ but this follows from Bernoulli's inequality: $$ \left(\frac{1+n}{2+n}\right)^{n+1}=\left(1-\frac{1}{2+n}\right)^{n+1}>1-\frac{n+1}{n+2}=\frac{1}{n+2}. $$
