Proof: Let $1 \leqslant \beta \leqslant n-1$, and $c$ is a finite decimal, then $c$ can be converted into a fraction, i.e., $c=\frac{a_{1}}{b_{1}}$, where $a_{1}, b_{1}$ are positive integers. When $1 \leqslant \beta \leqslant n-1$, then $\sqrt[n]{a}=b+c=\frac{b_{1} b+a_{1}}{b_{1}}$. That is, at this time $\sqrt[n]{a}$ can be expressed as a fraction, which contradicts Lemma 7, so $c$ cannot be a finite decimal but an infinite decimal. Suppose $c$ is a repeating decimal, then $c$ can also be converted into a fraction. Therefore, when $1 \leqslant \beta \leqslant n-1$, $\sqrt[n]{a}$ can also be expressed as a fraction, which contradicts Lemma 7. Hence, $c$ is not a repeating decimal, and the lemma is proved.