Prove $\sum_{n=0}^{\infty} \int_0^x t^n\sin \pi t dt$ converges uniformly in $[0,1]$.

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I probably solved it but I don’t know if it’s right. Following is my solution.

\begin{equation} \sum_{n=0}^{N-1} \int_0^x t^n\sin \pi t dt\leq \int_0^1\sum_{n=0}^{N-1}t^n \sin \pi tdt =\int_0^1\frac{1-t^N}{1-t}\sin \pi t dt\leq \int_0^1\frac{1}{1-t}\sin \pi t dt. \end{equation} Then let $1-t=x$, we get $$=\int_0^1\frac{\sin \pi(1-x)}{x}dx=\int_0^1\frac{\sin \pi x}{x}dx.$$ We know that this integral is convergent, so the original function term series converges uniformly. My question is whether the convergence of this integral and this problem will have a circular reasoning error? Maybe my question is stupid, because I don’t know which tools can be used and which can’t. Can anyone help me? Thank you!