I want to show that $({\log_2 x})^{n+1} \le x^n$ when $n \ge 1$ and $x \ge 1$.
I know that ${\log_2 x}$ can be shown to be $\lt x$ with:
$x \lt 2^x$
$\log_2 x \lt x$
and obviously adding the same exponent to either side would maintain this inequality but is there an equally straightforward way to prove that $({\log_2 x})^{n+1} \le x^n$ ?
$({\log_2 x})^{n+1} \le x^n$ for $n \ge 1$ and $x \ge 1$ is equivalent to $$ \frac {\ln x}{\ln 2} \le x^a $$ for $x \ge 1$, where $\ln$ is the natural logarithm, and $a = \frac{n}{n+1}$ is a real number in the range $\frac 12 \le a < 1$.
A simple analysis shows that the function $$ f(x) = \frac {\ln x}{\ln 2} - x^a $$ is increasing on $[1, x_0]$ and decreasing on $[x_0, \infty)$ where $$ x_0 = \frac{1}{(a \ln 2)^{1/a}} > 1 \quad . $$ So the desired inequality holds for all $x \ge 1$ if and only if $$ f(x_0) = \frac{-\ln a - \ln \ln 2 - 1}{a \ln 2} \le 0 \\ \Longleftrightarrow a \ge \frac{1}{e \ln 2} \approx 0.53074 \\ \Longleftrightarrow n \ge \frac{1}{e \ln 2 - 1 } \approx 1.131 \quad . $$ If you consider only positive integers $n$ then the inequality holds for all $x \ge 1$ exactly if $n \ge 2$.
For $n=1$, a counter-example is $x=8$: $ (\log_2 8)^2 = 3^2 = 9 > 8$.