3.1.15 ** Let $\left\{a_{n}\right\}$ be a geometric sequence with all terms being positive, and
$$
S=a_{1}+a_{2}+\cdots+a_{n}, T=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}},
$$

find the product of the first $n$ terms of $\left\{a_{n}\right\}$.
Let $\{a_{n}\}$ be a geometric sequence with common ratio $q$. Then, by the given conditions, we have
$$
S=a_{1}\left(1+q+q^{2}+\cdots+q^{n-1}\right), T=\frac{1}{a_{n}}\left(1+q+q^{2}+\cdots+q^{n-1}\right).
$$

Therefore, $a_{1} a_{n}=\frac{S}{T}$.
In a geometric sequence, the product of terms equidistant from the first and last terms is equal to the product of the first and last terms, i.e.,
$$
a_{1} a_{n}=a_{1} a_{n}, a_{2} a_{n-1}=a_{1} a_{n}, a_{3} a_{n-2}=a_{1} a_{n}, \cdots, a_{n} a_{1}=a_{1} a_{n},
$$

Multiplying the above $n$ equations, we get $\left(a_{1} a_{2} \cdots a_{n}\right)^{2}=\left(\frac{S}{T}\right)^{n}$, hence $a_{1} a_{2} \cdots a_{n}=\left(\frac{S}{T}\right)^{\frac{n}{2}}$.