What is the simplified value of $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)$?

Hint: An arithmetic progression can be given by $a,(a + d),(a + 2d),(a + 3d),...$ where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by$a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
Hence the given question is in the form of geometric progression.
For any geometric series given by $a{r^0},a{r^1},a{r^2},......,a{r^n}$
sum of the first $n$ terms of the geometric series are given by
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
Where,
$a$ is the first term of the geometric series,
$r$ is the ratio of the geometric series
$n$ is the number of terms in the series.
$r = \dfrac{{a{r^p}}}{{a{r^{p - 1}}}}$ for any $p \prec n$

Complete step-by-step solution:
We have to find simplified value $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)$. For that first, we are going to modify that to a geometric series.
By multiplying $\left( {2 + 1} \right)$ and $\left( {{2^2} + 1} \right)$ we will get,
$\left( {2 + 1} \right)\left( {{2^2} + 1} \right) = {2^3} + {2^2} + 2 + 1$
By multiplying $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)$and $\left( {{2^4} + 1} \right)$ we will get,
$\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right) = \left( {\left( {2 + 1} \right)\left( {{2^2} + 1} \right)} \right)\left( {{2^4} + 1} \right)$
We know the value of $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)$, by applying the value in the above equation we will get,
 $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right) = \left( {{2^3} + {2^2} + 2 + 1} \right)\left( {{2^4} + 1} \right)$
By multiplying the terms in the right hand side of the above equation we will get,
$\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right) = \sum\limits_{n = 0}^7 {{2^n}} $
By multiplying $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)$and $\left( {{2^8} + 1} \right)$ we will get,
$\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right) = \left( {\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)} \right)\left( {{2^8} + 1} \right)$
We know the value of $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)$ , by applying the value in the above equation we will get,
$\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right) = \left( {\sum\limits_{n = 0}^7 {{2^n}} } \right)\left( {{2^8} + 1} \right)$
By multiplying the two terms in the right hand side of the above equation, we will get
$\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right) = \sum\limits_{n = 0}^{15} {{2^n}} $.
Here, $\sum\limits_{n = 0}^{15} {{2^n}} $is a geometric series.
In this geometric series, from the hint we can get the values of $a$,$r$ and $n$.
For this above geometric series $\sum\limits_{n = 0}^{15} {{2^n}} $, we can write it as
$\sum\limits_{n = 0}^{15} {{2^n}} = \sum\limits_{n = 0}^n {a{r^n}} $.
If we put $n = 0$in the series, we can find the value of $a$.
Therefore, $a = {2^0}$
For all nonzero real numbers, let it be $x$, ${x^0} = 1$.
Therefore, $a = 1$
From the hint, let $p = 2$,
$r = \dfrac{{a{r^2}}}{{a{r^1}}}$
From the above geometric series,$a{r^2} = 4$ and $ar = 2$
Therefore,
$r = \dfrac{4}{2}$
$r = 2$.
Here we have to $16$ terms in this series, thus,
 $n = 16$
By applying these values in the formula, we will get,
$\sum\limits_{n = 0}^{15} {{2^n} = \dfrac{{1\left( {{2^{16}} - 1} \right)}}{{2 - 1}}} $
By doing some calculations we will get,
$\sum\limits_{n = 0}^{15} {{2^n} = {2^{16}} - 1} $.
Therefore, $\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)$$ = $${2^{16}} - 1$.

Note: Geometric Progression:
> In the GP the new series is obtained by multiplying the two consecutive terms so that they have constant factors.
> In GP the series is identified with the help of a common ratio between consecutive terms.
> Series vary in the exponential form because it increases by multiplying the terms.
The GM is known as the geometric mean which is the mean value or the central term in the set of numbers in the geometric progression.