One, (40 points) Find the smallest real number $\lambda$, such that there exists a sequence $\left\{a_{n}\right\}$ with all terms greater than 1, for which for any positive integer $n$ we have $\prod_{i=1}^{n+1} a_{i}<a_{n}^{\lambda}$.
Given $a_{n}>1$, so,
$$
\begin{array}{l}
\prod_{i=1}^{n+1} a_{i}0\right), S_{n}=\sum_{i=1}^{n} b_{i}\left(S_{n}>0\right) .
\end{array}
$$

For equation (1) to hold, then $\lambda>0$.
From equation (1) we get
$$
\begin{array}{l}
S_{n+2}S_{n+2}+\lambda S_{n} \geqslant 2 \sqrt{\lambda S_{n+2} S_{n}} \\
\Rightarrow \frac{S_{n+2}}{S_{n+1}}<\frac{\lambda}{4} \cdot \frac{S_{n+1}}{S_{n}} \\
\Rightarrow 1<\frac{S_{n+1}}{S_{n}}<\left(\frac{\lambda}{4}\right)^{n-1} \frac{S_{2}}{S_{1}} .
\end{array}
$$

Since the above inequality holds for any $n$, hence $\lambda \geqslant 4$. When $\lambda=4$, taking $a_{n}=10^{2 n}$ satisfies the conditions.
In conclusion, the minimum value of $\lambda$ is 4.