For the density $f(x\mid \theta)=\theta x^{\theta-1}$, is $\sum x_i$ sufficient for $\theta$?

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Suppose $x_1$, ..., $x_n$ be a random sample of distribution with pdf: $$f(x\mid \theta)=\theta x^{\theta-1}\mathbf1_{0<x<1},\,θ>0$$

Is $\sum_{i=1}^n x_i$ a sufficient statistic?

My classmate used the factorization theorem to prove that $\prod_{i=1}^nx_i$ is a sufficient statistic. Thus, $\sum_{i=1}^n x_i$ is not a sufficient statistic.

But I remembered that the sufficient statistics can be more than one. So to prove that $\sum_ix_i$ is not a sufficient statistic, we may need to prove that $\sum_ix_i$ can not be a function of the minimal sufficient statistic.

It seems that it is too complex to prove this. Proving by definition may be easier. But I don't know how to calculate $f(x\mid \sum _ix_i,\theta)$.

Am I wrong? Can we use the factorization theorem to prove? Will it be easy to prove by definition? If yes, then how to calculate $f(x\mid \sum _ix_i,\theta)$?

Thanks in advance!