Prove that $$ p(\sqrt[p]{n+1}-1)< \frac{1}{\sqrt[p]{1}}+\frac{1}{\sqrt[p]{2^{p-1}}}+...+\frac{1}{\sqrt[p]{n^{p-1}}}< p\sqrt[p]{n} \quad p\in \mathbb{N},p\ge 2 $$
I used AM-GM to prove it.
For right hand side of inequality
From AM-GM we get $n+1+(p-1)n>p\sqrt[p]{n^{p-1}(n+1)}$
$ \Longleftrightarrow $ $pn+1>p\sqrt[p]{n^{p-1}(n+1)}$
$ \Longleftrightarrow $ $1>p\sqrt[p]{n^{p-1}}(\sqrt[p]{n+1}-\sqrt[p]{n})$
$ \Longleftrightarrow $ $\frac{1}{\sqrt[p]{n^{p-1}}}>p(\sqrt[p]{n+1}-\sqrt[p]{n})$
$ \Longrightarrow $ $ p(\sqrt[p]{n+1}-1)<\frac{1}{\sqrt[p]{1}}+\frac{1}{\sqrt[p]{2^{p-1}}}+...+\frac{1}{\sqrt[p]{n^{p-1}}}$
For left hand side of inequality $n-1+(p-1)n>p\sqrt[p]{(n-1)n^{p-1}}$
$ \Longleftrightarrow $ $pn-1>p\sqrt[p]{(n-1)n^{p-1}}$
$ \Longleftrightarrow $ $\frac{1}{\sqrt[p]{n^{p-1}}}<p(\sqrt[p]{n}-\sqrt[p]{n-1})$
$ \Longrightarrow $ $\frac{1}{\sqrt[p]{1}}+\frac{1}{\sqrt[p]{2^{p-1}}}+...+\frac{1}{\sqrt[p]{n^{p-1}}}< p\sqrt[p]{n}$
Hope that someone will have nicer solution.
Note that $f(x)=\dfrac{1}{\sqrt[p]{x^{p-1}}}=x^{(1/p-1)}$ is a decreasing function
$$\frac{1}{\sqrt[p]{(k+1)^{p}}} <\int_{k}^{k+1} \frac{dx}{\sqrt[p]{x^{p-1}}} <\frac{1}{\sqrt[p]{k^{p}}}$$
$$\sum_{k=0}^{n-1} \frac{1}{\sqrt[p]{(k+1)^{p}}} <\int_{0}^{n} \frac{dx}{\sqrt[p]{x^{p-1}}} \implies \frac{1}{\sqrt[p]{1^{p}}}+\ldots+\frac{1}{\sqrt[p]{n^{p}}} < p\sqrt[p]{n}$$
$$\int_{1}^{n} \frac{dx}{\sqrt[p]{x^{p-1}}} < \sum_{k=1}^{n} \frac{1}{\sqrt[p]{k^{p}}} \implies p\left( \sqrt[p]{n+1}-1 \right)<\frac{1}{\sqrt[p]{1^{p}}}+\ldots+\frac{1}{\sqrt[p]{n^{p}}}$$