Proof of the inequality $(p+1)n^p<(n+1)^{p+1}-n^{p+1}<(p+1)(n+1)^p$ Question:
Let $n, p \in \mathbb{N}$. Prove that
$$(p+1)n^p<(n+1)^{p+1}-n^{p+1}<(p+1)(n+1)^p \tag{$\star$}$$

What I have noticed so far is, that with $f(x) := x^p$ I can conclude that
$$\int_{n}^{n+1}f(x)\,dx = \int_{n}^{n+1}x^p\,dx = \frac{1}{p+1}(n+1)^{p+1}-\frac{1}{p+1}n^{p+1}$$
Thus $\star$ is correct iff
$$n^p<\int_{n}^{n+1}x^p\,dx<(n+1)^p \tag{$\star\star$}$$
Which is the same as
$$f(n)<\int_{n}^{n+1}f(x)\,dx<f(n+1)$$

This looks not so hard, but anyhow, here I get stuck and I do not know how to show this inequality.
 A: Use the fact that $f(x)$ is increasing, and so $f(n) \leq f(x) \leq f(n+1)$, for all $x \in [n,n+1].$ Now, integrate all the terms. $f(n)$ and f(n+1) being constants, the integral over an interval of length 1 is equal to that constant. Then what you have written gives you the solution. Also, you would have strict inequality, as $f$ is strictly increasing.
A: The easiest way is by means of the mean/intermediate value theorem.  Note that
$(n+1)^{p+1} - n^{p+1} = (p+1) (n^*)^p$
for some value $n^*$ between $n$ and $n+1$.  This establishes the inequality for all positive reals $n$ and hence automatically also for the integers.
A: A very simple approach is based on binomial expansion $$(n + 1)^{p + 1} = n^{p + 1} + (p + 1)n^{p} + \text{ some positive terms}$$ which immediately gives the inequality $$(n + 1)^{p + 1} - n^{p + 1} > (p + 1)n^{p}$$
Next we can see that
$\displaystyle \begin{aligned}(n + 1)^{p + 1} - n^{p + 1}&= (p + 1)n^{p} + \binom{p + 1}{2}n^{p - 1} + \cdots\\
&= (p + 1)\left\{n^{p} + \frac{p}{2!}n^{p - 1} + \frac{p(p - 1)}{3!}n^{p - 2} + \cdots\right\}\\
&< (p + 1)\left\{n^{p} + \frac{p}{1!}n^{p - 1} + \frac{p(p - 1)}{2!}n^{p - 2} + \cdots\right\}\\
&= (p + 1)(n + 1)^{p}\end{aligned}$
Inequalities involving rational numbers and algebraical functions are normally provable using Algebra. Calculus is needed when we wish to extend these to irrational numbers or when we deal with transcendental functions.
A: for the second inequality one nay note that:
$$
(n+1)^{p+1} - n^{p+1} = [(n+1)-n][(n+1)^pn^0 + (n+1)^{p-1}n^1+\dots+(n+1)^0n^p] \\
\lt [1][(n+1)^p+\dots +(n+1)^p]=(p+1)(n+1)^p
$$
