First, prove that if $a$ is not divisible by any integer $>1$ and $\leqslant \sqrt{a}$, then $a$ is a prime number. Assume $a$ is a composite number and $a=b c$, where $b$ and $c$ are both integers greater than 1. Since $a$ is not divisible by any integer $>1$ and $\leqslant \sqrt{a}$, it follows that $b>\sqrt{a}$ and $c>\sqrt{a}$, and thus $b c>\sqrt{a} \cdot \sqrt{a}=a$, which contradicts $b c=a$. Therefore, if $a$ is not divisible by any integer $>1$ and $\leqslant \sqrt{a}$, then $a$ is a prime number.

From the above, if $a$ is a composite number, then $a$ must have a divisor $>1$ and $\leqslant \sqrt{a}$. By Lemma 5, the smallest divisor of $a$ greater than 1 must be a prime number, hence the lemma is proved.

Assume $n \geqslant 2$ is an integer, define
$$a_{1} a_{2} \cdots a_{n}=\left\{\begin{array}{ll} a_{1} a_{2}, & \text { when } n=2 \text {; } \\ a_{1} a_{2} a_{3}, & \text { when } n=3 \text {; } \\ a_{1} a_{2} a_{3} a_{4}, & \text { when } n=4 \text {; } \\ a_{1} a_{2} a_{3} a_{4} \cdots a_{n}, & \text { when } n \geqslant 5 \text {; } \end{array}\right.$$