Let \( n \geq 2 \) be some fixed positive integer and suppose that \( a_{1}, a_{2}, \ldots, a_{n} \) are positive real numbers satisfying \( a_{1}+a_{2}+\cdots+a_{n}=2^{n}-1 \).

Find the minimum possible value of

\[
\frac{a_{1}}{1}+\frac{a_{2}}{1+a_{1}}+\frac{a_{3}}{1+a_{1}+a_{2}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}} .
\]
We claim the the minimum possible value of this expression is $n$. Observe that by AM-GM, we have that

$$
\begin{aligned}
\frac{a_{1}}{1}+ & \frac{a_{2}}{1+a_{1}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}} \\
& =\frac{1+a_{1}}{1}+\frac{1+a_{1}+a_{2}}{1+a_{1}}+\cdots+\frac{1+a_{1}+a_{2}+\cdots+a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}-n \\
& \geq n \cdot \sqrt[n]{\frac{1+a_{1}}{1} \cdot \frac{1+a_{1}+a_{2}}{1+a_{1}} \cdots \frac{1+a_{1}+a_{2}+\cdots+a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}}-n \\
& =n \cdot \sqrt[n]{1+a_{1}+a_{2}+\cdots+a_{n}}-n \\
& =2 n-n=n .
\end{aligned}
$$

Furthermore, equality is achieved when $a_{k}=2^{k-1}$ for each $1 \leq k \leq n$.

Note: The text provided is already in English, so no translation was necessary.