I understand it should be so, considering the definition:
$L^{\infty}( \Omega)= \{ f: \Omega \to \mathbb{R}\, \mid f$ is measurable and there is $C \in \mathbb{R^{+}}$ such that $|f(x)| \leq C$ a.e. on $\Omega \}$
That is, provided $f \in L^{\infty}(\Omega)$, we need to show that $f^p$ defined as $f^p= (f(x))^p$, belongs to $L^{\infty}(\Omega)$:
We have $|f(x)| \leq C$ almost everywhere on $\Omega$. So, as for $1<p<\infty$ the function $x^p$ is non decreasing for $x$ positive, we have $(f(x))^p \leq C^p$ almost everywhere.
I think that is pretty trivial, but I have a little confusion as we are not dealing with plain supremum, but with the essential supremum, so can we ensure the inequality indeed holds almost everywhere?
As pointed out in the comments, if $\left|f(x)\right|\leqslant C$ for almost every $x$, then using increasingness of the map $t\mapsto t^p$ for non-negative $t$, we have $\left|f(x)\right|^p\leqslant C^p$ hence $\left|f\right|^p$ satisfies the definition of $L^\infty(\Omega)$.