Example 5 The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=10, a_{n+1}=\sqrt[n]{a_{n}}(n \geqslant 1)$. Find $a_{n}$.
Solution: Since $a_{1}=10>0$, we have
$$
a_{2}>0, a_{3}>0, \cdots \cdots a_{n+1}>0 .
$$

Taking the common logarithm of both sides of $a_{n+1}=\sqrt[n]{a_{n}}$ yields $\lg a_{n+1}=\frac{1}{n} \lg a_{n}$.
Let $b_{n}=\lg a_{n}$. Then $b_{n+1}=\lg a_{n+1}$.
Thus, $b_{n+1}=\frac{1}{n} b_{n}$, which means $\frac{b_{n+1}}{b_{n}}=\frac{1}{n}$.
Noting that $b_{1}=\lg a_{1}=1$, we have
$$
\begin{array}{l}
b_{n}=\frac{b_{n}}{b_{n-1}} \cdot \frac{b_{n-1}}{b_{n-2}} \cdots \cdots \frac{b_{2}}{b_{1}} \cdot b_{1} \\
=\frac{1}{n-1} \cdot \frac{1}{n-2} \cdots \cdots \frac{1}{2} \cdot \frac{1}{1} \cdot 1 \\
=\frac{1}{(n-1)!} .
\end{array}
$$

Therefore, $\lg a_{n}=\frac{1}{(n-1)!}$, which means $a_{n}=10^{(n-1)!}$.