<s>[INST] Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \frac{\sum_{k=0}^{p}(n+1)^{p-k}n^{k}}{p+1} = \frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1}$$
I'm not sure where to proceed from here. Am I supposed to use a geometric series in here somewhere?
EDIT: I may have figured it out if someone can confirm that $(n+1)^{p-k}n^{k} > n^{p}$,  $1 \le k \le p$, $\forall n$
Assuming the above is true, we have:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \gt \frac{n^{p} + n^{p} + .... + n^{p}  + n^{p}}{p+1} = \frac{(p+1)n^{p}}{(p+1)} = n^{p}$$
Similarly:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \lt \frac{(n+1)^{p} + (n+1)^{p}  + .... + (n+1)^{p} + (n+1)^{p}}{p+1} = \frac{(p+1)(n+1)^{p}}{(p+1)} = (n+1)^{p}$$
 [/INST] Use the mean value theorem. ${}{}{}{}{}{}{}$
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