15. Since $a_{1}=2$, and for all positive integers $n$ we have $a_{n+1}=a_{1} a_{2} \cdots a_{n}+1$, therefore, for all positive integers $n$ we have $a_{n} \geqslant 2, a_{1} a_{2} \cdots a_{n} \geqslant 2^{n}$.
. We will prove by mathematical induction that for all positive integers $n$,
$$1-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)=\frac{1}{a_{1} a_{2} \cdots a_{n}} .$$

When $n=1$, by $a_{1}=2$, we have $1-\frac{1}{a_{1}}=\frac{1}{a_{1}}$, so the conclusion holds.

Assume that for $n=k(k>1, k \in \mathbf{N})$, the conclusion holds, i.e., $1-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}\right)=\frac{1}{a_{1} a_{2} \cdots a_{k}}$.
Then, by $a_{k+1}=a_{1} a_{2} \cdots a_{k}+1$, we have
$1-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}+\frac{1}{a_{k+1}}\right)$
$=\frac{1}{a_{1} a_{2} \cdots a_{k}}-\frac{1}{a_{k+1}}=\frac{a_{k+1}-a_{1} a_{2} \cdots a_{k}}{a_{1} a_{2} \cdots a_{k} a_{k+1}}$
$=\frac{1}{a_{1} a_{2} \cdots a_{k} a_{k+1}}$.
Thus, when $n=k+1$, the conclusion also holds.
In summary, for any positive integer $n$ we have
$1-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)=\frac{1}{a_{1} a_{2} \cdots a_{n}}$.
Combining this with $a_{1} a_{2} \cdots a_{n} \geqslant 2^{n}$, we know that for all positive integers $n$,
$1-\frac{1}{a_{1}}-\frac{1}{a_{2}}-\cdots-\frac{1}{a_{n}}=\frac{1}{a_{1} a_{2} \cdots a_{n}} \leqslant \frac{1}{2^{n}}$.
Thus, $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \geqslant 1-\frac{1}{2^{n}}$.
Notice that
$$\begin{array}{l} \frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n}}=\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}=1-\frac{1}{2^{n}} . \\ \text { Hence } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \geqslant \frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n}} \end{array}$$

For all positive integers $n$, the above inequality holds.