So, this is how I solved
$\displaystyle T(n-1) \approx{} T(n-2) $
$\displaystyle T(n) = T(n-1)^2 $
Add log in both sides
$\displaystyle log(T(n)) = 2log(T(n-1)) $
$\displaystyle Let\:\: log(T(n)) = S(n) $
$\displaystyle S(n) = 2(S(n-1)) $
We know that from Fibonacci series
$\displaystyle S(n) = O(2^n) $
$\displaystyle log(T(n)) = O(2^n) $
$\displaystyle (T(n)) = O(2^{2^n}) $
Ok, so where did I go wrong here??
So, this is how I solved
$T(n-1) \approx{} T(n-2)$
$T(n) = T(n-1)^2$
An equality is the most precise and most powerful entity in the world. Please treat it with due respect.
Let me show how wrong it is if you use approximation liberally with equalities. If your argument above is sound, we can also do the following.
$\quad T(n) \approx{} T(n-1)$
$\quad T(n-1) = T(n-1)*T(n-2)$
$\quad T(n-2) = 1$ if $T(n-1)$ is not 0.
So $T(n)$ must be 0 or 1 from time to time . That is obviously false, however. For example, we can let $T(1)=T(2)=2$.
What you can do is, as gnasher729 suggested, "taking logarithm on both sides immediately", assuming $T(1)>0$ and $T(2)>0$, which can be assume generally when we are analyzing recurrence relations in computer science.
You can also, of course, dive into analysis by cases when $T(1)$ and $T(2)$ might be 0 or even negative.
