$\left(\int_a^b f(x)^n dx\right)^{1/n}$ converges to $\sup\{f(x):x\in [a,b]\}$.

Question

Let $I$ be the interval $[a,b]$ and suppose that $f:I\to\mathbb{R}$ is continuous on $I$ with $f(x)\geq 0$ $\forall x\in I$. How can we show that the sequence $$\left(\int_a^b f(x)^n dx\right)^{1/n}$$ for $n\in\mathbb{N}$ converges to $M:=\sup\{f(x):x\in I\}$?