How to find the nth term of the sequence 2 , 4 , 16 ,...

Okay, we have ${\displaystyle 2,4,16,256,....?}$
Look for a shared distinction between each. They are all divisible by ${\displaystyle 2}$. Since they are all divisible by ${\displaystyle 2}$, we have ${\displaystyle 2^1,2^2,2^4,2^8,...?}$
Now let's look at the power exponents, ${\displaystyle 1,2,4,8,...?}$
It looks like for ${\displaystyle 1,2,4,8,...?}$ can work if we have ${\displaystyle 2^n}$, starting a ${\displaystyle 0}$.
Now we have ${\displaystyle 2^{2^n}}$
Plug in to be sure:
${\displaystyle 2^{2^0}=2}$
${\displaystyle 2^{2^1}=4}$
${\displaystyle 2^{2^3}=16}$
${\displaystyle 2^{2^4}=256}$

Given:

${\displaystyle 2,4,16,256,...}$

Each element of the sequence appears to be the square of the one before it, as shown by:

${\displaystyle 4=2^2}$
${\displaystyle 16=4^2}$
${\displaystyle 256={16}^2}$

This would result in the formula:

${\displaystyle a_n=2^{2^n}}$

However, note that we have been told nothing about the nature of this sequence except the first ${\displaystyle 4}$ terms. We have not even been told that it is a sequence of numbers.
For example, it can be matched with a cubic formula:

${\displaystyle a_n=\frac{1}{3}(109n^3-639n^2+1160n-624)}$

Then it would not follow the squaring pattern, but would continue:

${\displaystyle 2,4,16,256,942,2292,4524,...}$

We could choose any following numbers we like and find a formula that matches them.
No infinite sequence is determined purely by its first few terms.