I know that in quadratic sequence the second common difference is constant, so I want to apply this idea to geometric sequences, like the second ratio is constant, example :
$$ 1, 2, 8 , 64 , 1024 ... $$
so it's : $$ ×2, ×4, ×8, ×16... $$
which is : $$ ×2, ×2, ×2,... $$
So how do we find the $nth$ term here?
Bonus :
How do we find the $nth$ term in this type of sequence :
$$ 1,2,6,24,120,...$$
so it's :
$$ ×2,×3,×4,...$$
which is :
$$+1,+1,+1,+1,...$$
The first sequence is $2^N$ where the exponent $N$ is the sum of the geometric series $1+2 + \cdots + k$. That sum is well known. You can finish the details.
The second sequence is just the sequence of factorials $n!$.
Whenever you see a product of terms, it’s worth asking about what happens if you take the logarithm of the product. The product rule of logarithms allows you to turn a log of a product into a sum of logarithms, which might make a pattern easier to spot.
In your case, since you keep multiplying by 2, let’s try taking the base-2 logarithm of each term in the sequence. That gives back the sequence
$$0, 1, 3, 6, 10, 15, \dots$$
Do you recognize this sequence? If so, you can get a nice formula for it, then undo the logarithm by raising 2 to that power to get the solution.
The sequence 1,2,8,64,1024 has a closed form in terms of $n$. We first write each term as a power of $2$ to search for some pattern, $u_0 = 2^0, u_1=2^1, u_2=2^3, u_3=2^6, u_4 = 2^{10}$. We can then simply conclude that at $n$, the sequence can be written as $u_n = 2^{\frac{n(n+1)}{2}}$.This is of course by pattern.
Another way to look at it, is at each stage $k$, we are multiplying $u_{k-1}$ by $2^k$, so $\frac{u_n}{u_{n-1}} = 2^n$. Now we will consider the following product: $$ \prod_{k=1}^n\frac{u_k}{u_{k - 1}} = \frac{u_{n}}{u_{n -1}}\times...\times\frac{u_{2}}{u_1}\times\frac{u_{1}}{u_0} = \frac{u_n}{u_0} = \prod_{k=1}^n 2^k = 2^{\frac{n(n+1)}{2}}\Rightarrow $$ $$ \boxed{u_n = 2^{\frac{n(n+1)}{2}}} $$
Like what @ethan-bolker said, it's just the sequence $u_n = n!$.
