Is this geometric mean-like limit equal $0$?

Let $$(a_n)_n$$ be a sequence of positive real numbers such that $$a_n\leq1$$ for all $$n\in\mathbb{N}$$, and suppose that $$\displaystyle\lim_{n\to\infty}a_n=0$$. By the Stolz–Cesàro theorem we know that $$\displaystyle\lim_{n\to\infty}\left[\prod_{n=1}^na_n\right]^\frac{1}{n}=0$$ $$\quad(\star)$$.

For each $$n\in\mathbb{N}$$, let $$S_n$$ be a random subset of $$\{1,2,\dots,n\}$$ and let $$c_n$$ be the cardinality of $$S_n$$. Put $$b_n=\left[a_{n+1}\left(\prod_{k\in S_n}a_k\right)\right]^\frac{1}{1+c_n}.$$

Question: Is $$\displaystyle\lim_{n\to\infty}b_n=0$$?

If I try to imagine cases, I only can think of $$\displaystyle\lim_{n\to\infty}b_n=0$$, but I don't know how to prove this.

For example, if $$S_n\subset S_{n+1}$$ (proper subset) for all $$n\in\mathbb{N}$$, then the factors that define $$b_n$$ form a sequence that converges to zero (subsequence of $$(a_n)_n$$), and therefore $$\displaystyle\lim_{n\to\infty}b_n=0$$ by $$(\star)$$.

Any insight for the general case would be appreciated.

Fix $$\varepsilon \in (0, 1)$$. Note that there are only finitely many $$a_n$$ such that $$a_n \ge \varepsilon$$; let $$K$$ be the number of such sequence terms (note that this depends on $$\varepsilon$$, not $$n$$). Since $$a_n \to 0$$, there exists some $$N$$ such that $$n > N \implies a_n < \varepsilon^{K+1}.$$
For each $$n$$, define \begin{align*} U_n &= \{n \in S_n : a_n \ge \varepsilon\} \\ L_n &= \{n \in S_n : a_n < \varepsilon\} = S_n \setminus U_n. \end{align*}
Please note that $$|U_n| \le K$$ for all $$n$$. Then, \begin{align*} n > N &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{K+1} \prod_{k\in S_n}a_k\\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{K+1} \prod_{k\in U_n}a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{|U_n| + 1} \prod_{k\in U_n}a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon \prod_{k\in U_n}\varepsilon a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon \prod_{k\in U_n}\varepsilon \prod_{k\in L_n}\varepsilon \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{|S_n|+1} \\ &\implies \left(a_{n+1}\prod_{k\in S_n}a_k\right)^{\frac{1}{|S_n|+1}} < \varepsilon. \end{align*}