Prove that because $\sum_{i=1}^{m} \frac{1}{a_{i}}=1 \geqslant m \sqrt[m]{\frac{1}{a_{1} a_{2} \cdots a_{m}}}$, so
$$a_{1} a_{2} \cdots a_{m} \geqslant m^{m}$$

According to the polynomial expansion theorem, we have
$$\left(\sum_{i=1}^{m} a_{i}\right)^{n}=\sum \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{1}^{n_{1}} a_{2}^{n_{2}} \cdots a_{m}^{n_{m}}$$

where $n_{1}, n_{2}, \cdots, n_{m}$ are non-negative integers, and $n_{1}+n_{2}+\cdots+n_{m}=n$. Thus,
$$\left(\sum_{i=1}^{m} a_{i}\right)^{n}-\sum_{i=1}^{m} a_{i}^{n}=\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{1}^{n_{1}} a_{2}^{n_{2}} \cdots a_{m}^{n_{m}}$$

where $n_{1}, n_{2}, \cdots, n_{m}$ are non-negative integers, and less than $n, n_{1}+n_{2}+\cdots+n_{m}=n$ (the following $\sum^{\prime}$ all satisfy this condition).

In (1), replace the letters $a_{i}$ with 1, we get
$$\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!}=m^{n}-m$$

By symmetry, we have
$$\begin{aligned} & \left(\sum_{i=1}^{m} a_{i}\right)^{n}-\sum_{i=1}^{m} a_{i}^{n}=\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{2}^{n_{1}} a_{3}^{n_{2}} \cdots a_{1}^{n} \\ = & \cdots=\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{m}^{n_{1}} a_{1}^{n_{2}} \cdots a_{m-2}^{n_{m-1}-1} a_{m-1}^{n_{m}} \end{aligned}$$

Multiply (1) with the above $n-1$ equalities, and apply the generalized Cauchy inequality, we get
$$\begin{aligned} & {\left[\left(\sum_{i=1}^{m} a_{i}\right)^{n}-\sum_{i=1}^{m} a_{i}^{n}\right]^{m} } \\ = & \sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{1}^{n_{1}} a_{2}^{n_{2}} \cdots a_{m}^{n} \sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} \cdot \\ & a_{2}^{n_{1}} a_{3}^{n_{2}} \cdots a_{1}^{n} \cdots \sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{m}^{n_{m}} a_{1}^{n_{2}} \cdots a_{m-2}^{n} a_{m-1}^{n} \\ \geqslant & {\left[\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!}\left(a_{1} a_{2} \cdots a_{m-1} a_{m}\right)^{\frac{n_{1}}{m}}\left(a_{2} a_{3} \cdots a_{m} a_{1}\right)^{\frac{v_{2}}{m} \cdots}\right.} \\ & \left.\left(a_{m} a_{1} a_{2} \cdots a_{m-1}\right)^{\frac{n_{m}}{m}}\right]^{m} \\ = & {\left[\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!}\left(a_{1} a_{2} \cdots a_{m-1} a_{m}\right)^{\frac{n_{1}+n_{2}+\cdots+n_{m}}{m}}\right]^{m} } \\ = & {\left[\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!}\left(a_{1} a_{2} \cdots a_{m-1} a_{m}\right)^{\frac{n}{m}}\right]^{m} } \\ = & \left(a_{1} a_{2} \cdots a_{m-1} a_{m}\right)^{n}\left(\sum^{\prime} \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!}\right)^{m} \\ \geqslant & m^{m}\left(m^{n}-m\right)^{m}=\left(m^{2 n}-m^{n+1}\right)^{m}, \end{aligned}$$

Taking the $m$-th root on both sides, we get
$$\left(\sum_{i=1}^{m} a_{i}\right)^{n}-\sum_{i=1}^{m} a_{i}^{n} \geqslant m^{2 n}-m^{n+1}$$