Solving the recurrence $T(n)=T(n-1)*T(n-2)$
Your reasoning is perfect. To wit, here is what you get if you don't make any mistakes in your algebra. As you mention, $S(n)$ satisfies the Fibonacci recurrence, and furthermore $S(1) = 2 = F_3$, $S(0) = 1 = F_2$. We conclude that $S(n) = F_{n+2}$. Therefore $T(n) = 2^{F_{n+2}}$. The first few values are
$$
2, 4, 8, 32, 256, \ldots
$$
To get a closed formula, note that
$$
F_n = \frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^n.
$$
Therefore
$$
\begin{align*}
S(n) &amp;= \frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^{n+2} - \frac{1}{\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^{n+2} \\ &amp;=
\frac{3+\sqrt{5}}{2\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{3-\sqrt{5}}{2\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^n \\ &amp;=
\frac{5+3\sqrt{5}}{10} \left(\frac{1+\sqrt{5}}{2}\right)^n + \frac{5-3\sqrt{5}}{10} \left(\frac{1-\sqrt{5}}{2}\right)^n,
\end{align*}
$$
using the fact that $(1\pm\sqrt{5})/2$ are roots of $x^2=x+1$.