3. Given an integer $n \geqslant 2$. Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, a_{1}+a_{2}+\cdots+a_{n}=n .
$$

Find the minimum value of $a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\cdots+a_{1} a_{2} \cdots a_{n}$.
For an integer $m \geqslant 1$, let $f_{m}\left(x_{1}, x_{2}, \cdots, x_{m}\right)=x_{1}+x_{1} x_{2}+x_{1} x_{2} x_{3}+\cdots+x_{1} x_{2} \cdots x_{m}$.
Lemma: If $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{m} \geqslant 0$, and the average of $x_{1}, x_{2}, \cdots, x_{m}$ is $A$, and $A \leqslant 1$,
then $f_{m}\left(x_{1}, x_{2}, \cdots, x_{m}\right) \geqslant f_{m}(A, A, \cdots, A)$.
We use mathematical induction. When $m=1$, the conclusion is obviously true.
Assume the conclusion holds for $m$, consider the case for $m+1$. Let the average of $x_{2}, \cdots, x_{m}, x_{m+1}$ be $B$, by the problem, $B \leqslant A \leqslant 1$, and $x_{1}+m B=(m+1) A \Rightarrow x_{1}=(m+1) A-m B$.
$$
\begin{array}{l}
\text { Then } f_{m+1}\left(x_{1}, x_{2}, \cdots, x_{m}, x_{m+1}\right)=x_{1}\left(1+x_{2}+x_{2} x_{3}+\cdots+x_{2} \cdots x_{m} x_{m+1}\right) \\
=x_{1}\left[1+f_{m}\left(x_{2}, \cdots, x_{m}, x_{m+1}\right)\right] \geqslant x_{1}\left[1+f_{m}(B, B, \cdots, B)\right] \\
=[(m+1) A-m B]\left(1+B+B^{2}+\cdots+B^{m}\right)=m(A-B) \sum_{i=0}^{m} B^{i}+A \sum_{i=0}^{m} B^{i} . \\
\text { Thus } f_{m+1}\left(x_{1}, x_{2}, \cdots, x_{m}, x_{m+1}\right)-f_{m+1}(A, A, \cdots, A, A) \\
=m(A-B) \sum_{i=0}^{m} B^{i}+A \sum_{i=0}^{m} B^{i}-\sum_{i=1}^{m+1} A^{i}=m(A-B) \sum_{i=0}^{m} B^{i}-A \sum_{i=0}^{m}\left(A^{i}-B^{i}\right) \\
=m(A-B) \sum_{i=0}^{m} B^{i}-A(A-B)\left[1+\sum_{i=2}^{m}\left(A^{i-1}+A^{i-2} B+\cdots+A B^{i-2}+B^{i-1}\right)\right] \\
\geqslant m(A-B) \sum_{i=0}^{m} B^{i}-(A-B)\left[1+\sum_{i=2}^{m}\left(1+B+\cdots+B^{i-2}+B^{i-1}\right)\right] \\
=m(A-B) \sum_{i=0}^{m} B^{i}-(A-B)\left[m+(m-1) B+(m-2) B^{2}+\cdots+B^{m-1}\right] \\
=(A-B)\left(B+2 B+\cdots+m B^{m}\right) \geqslant 0, \text { i.e., the conclusion also holds for } m+1.
\end{array}
$$

By the principle of mathematical induction, the lemma is proved.
Returning to the original problem, by the lemma, when $a_{1}=a_{2}=\cdots=a_{n}=1$, the minimum value of $a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\cdots+a_{1} a_{2} \cdots a_{n}$ is $n$.