For what values of $k$ is there a perfect $x^p$ in {$n, n+1, n+2, ... ,kn$}? I was thinking about this the other day, and I can't seem to find the answer.
It is a fairly simple proof by induction to show that for all $n ∈ ℕ$, there is a perfect square in ${n, n+1, n+2, ... , 2n}$.
I am trying to generalize this question. For what values of $k$ is there always perfect cube in ${{n, n+1, n+2,..., kn}}$?
More generally, for what values of $k$ is there always a perfect $x^p$ in ${n, n+1, n+2, ... , kn}$? I'm not sure how to approach a proof of this. Maybe it's way above me?
Thanks.

For $n = 1$, since $1$ is a $p$'th power, any $k$ will do. For $n = 2$, the next largest $p$'th power if $2^p$, so
$$kn \ge 2^p \implies k \ge 2^{p-1} \tag{1}\label{eq1A}$$
I will show the minimum allowed value of $k = 2^{p-1}$ always works. For any $2 \le n \le 2^p$, this value of $k$ works. Also, if $n = m^p$ for any integer $m$, then this $k$ also works. Next, consider
$$m^p \lt n \lt (m+1)^p \tag{2}\label{eq2A}$$
for an integer $m \ge 2$.
For $p = 1$, we have $k = 1$, with this working since each value is it's own first power. For $p = 2$, we have $k = 2$, which you've stated you can prove using induction. Thus, consider $p \ge 3$. Using \eqref{eq2A}, the Binomial theorem expansion with $x \gt 0$ giving $(1 + x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \ldots + x^p \gt 1 + px$, and $m \ge 2 \implies m + m \ge m + 2 \implies 2m - 1 \ge m + 1$, we then get
$$\begin{equation}\begin{aligned}
\left(2^{p-1}\right)n & \gt \left(2^{p-1}\right)m^p \\
& = \left(\frac{1}{2}\right)\left(2^{p}\right)m^p\left(\frac{(m+1)^p}{(m+1)^p}\right) \\
& = \left(\frac{1}{2}\right)\left(\frac{2m}{m+1}\right)^p(m+1)^p \\
& = \left(\frac{1}{2}\right)\left(1 + \frac{m - 1}{m+1}\right)^p(m+1)^p \\
& \gt \left(\frac{1}{2}\right)\left(1 + \frac{p(m - 1)}{m+1}\right)(m+1)^p \\
& \ge \left(\frac{1}{2}\right)\left(1 + \frac{3(m - 1)}{m+1}\right)(m+1)^p \\
& = \left(\frac{1}{2}\right)\left(\frac{4m - 2}{m+1}\right)(m+1)^p \\
& = \left(\frac{2m - 1}{m+1}\right)(m+1)^p \\
& \ge (m+1)^p
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, $(m+1)^p \in [n,kn]$, confirming $k = 2^{p-1}$ (as well as any larger $k$) will always work.
