Clearly, no $S_n$ will equal 0. Hence dividing both sides of the given recursion by $S_{n-1}^2S_{n-2}$ leads to $$\Big(\frac{S_n}{S_{n-1}}\Big)^2=10\times\Big(\frac{S_{n-1}}{S_{n-2}}\Big)$$ Let $T_n = \frac{S_n}{S_{n-1}}$, $n\ge 3$. Then $T_3=\frac{S_3}{S_2}=\frac{0+1+9}{0+1}=10$ and \begin{align*} T_n & =10^{\frac{1}{2}}\times T_{n-1}^{\frac{1}{2}}\\ T_{n-1}^{\frac{1}{2}} &= 10^{\frac{1}{2^2}}\times T_{n-2}^{\frac{1}{2^2}}\\ T_{n-2}^{\frac{1}{2^2}} &= 10^{\frac{1}{2^3}}\times T_{n-3}^{\frac{1}{2^3}}\\ \cdots \\ T_4^{\frac{1}{2^{n+4}}} &= 10^{\frac{1}{2^{n+3}}}\times T_3^{\frac{1}{2^{n+3}}} \end{align*} Multiplying these relations and noticing $T_3=10$ yields: $$T_n = 10^{\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n+3}}}\times 10^{\frac{1}{2^{n+3}}}=10$$ This means $S_{n}/S_{n-1}=T_n = 10$ is a constant, or $\{T_n\}$ is a geometric sequence whose common ratio is 10 and $T_3=10$. Hence, by \myJustRefP{eq_geometric_seq_an1}, we have $$S_n =10\times 10^{n-3}=10^{n-2}\qquad(n\ge 3)$$ Finally by \myJustRefP{eq_seq_sn_an}, we have $$a_n = S_n - S_{n-1} = 10^{n-2}-10^{n-3}=\boxed{9\times 10^{n-3}}\quad (n \ge 3)$$