First, I would want to point out that $\lim_{n\rightarrow\infty} x^n=0$ on the interval $[0,1)$, thus $x^nf(x)\rightarrow 0$ (pointwise).
I believe this is a good use of the dominated convergence theorem, as $f\in L^1$, and for all $n$, $x^n f(x)\leq f(x)$ on our interval $[0,1)$, thus DCT says $\lim x^nf(x)\in L^1$, and even more so, $\lim\int x^nf(x) dx=\int\lim x^nf(x)dx=\int 0=0$.
However, I have not necessarily said that each $x^nf(x)\in L^1$ on their own yet, so I will use Holder's inequality to say, for each $n$: $|x^nf(x)|\leq |x^n|_\infty |f(x)|<\infty$ as $f(x)\in L^1$ and $|x^n|_\infty$ is finite.
My question still then is, is $|x^n|_\infty$ finite? I certainly believe it should be, I think its even 0, but I am not sure how to show that. Is it just a matter of $\int (x^n)^\infty dx=0$? That feels cheap. I suppose the issue here is more so a lack of my own understanding of the essential supremum.