Simplify the expression $(a+1)(a^2+1)(a^4+1)\cdots(a^{32}+1)$ How do I simplify this expression?

$$\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$$

 A: $$=\sum_{k=0}^{2m-1}a^{k}=\sum_{k=0}^{63}a^{k}$$
for $a\neq 1$ (geometric series) see https://en.wikipedia.org/wiki/Geometric_series
$$\sum_{k=0}^{2m-1}a^{k}=\frac{1-a^{2m}}{1-a}$$
A: Hint:
Multiply and divide by $a-1$, and prove by an easy induction that
$$\prod_{k=0}^n\bigl(a^{2^k}+1\bigr)=\frac{a^{2^{n+1}}-1}{a-1}.$$
A: Oh this is so cute!
We know that $(x + y)(x - y) = x^2 - y^2$ and therefore for for any $(a^n - 1)(a^n + 1) = (a^{2n} - 1)$.
So if we simply multiply $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$ by $(a-1)$ we get
$(a -1)\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$(a^2 - 1)\left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
And the whole thing collapses like dominoes:
$\left( a^{4}-1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$\left( a^{8}-1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$ \left( a^{16}-1 \right)\left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$ \left( a^{32}-1 \right) \left( a^{32}+1 \right)=$
$ \left( a^{64}-1 \right)$
So $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)  = \frac {a^{64} - 1}{a - 1}$.
But here's are two other another interesting observations:
$(a - 1)(a^n + a^{n-1} + ...+a + 1) = (a^{n+1} + a^n + ... + a^2 + a) - (a^n + a^{n-1} +... + a + 1) = (a^{n+1} - 1)$
So $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right) = \frac {a^{64} - 1}{a - 1} = (a^{63} + a^{62} + .... + a^2 + a + 1)$
Which we can verify directly by noting
$(a^n + 1)( a^{n-1} + ...+ a + 1) = a^{2n - 1} + a^{2n - 2} + ... + a + 1$
so  $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$(a^3 + a^2 + a + 1) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
$(a^7 + a^6+.. + a + 1)  \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$
...
$(a^{63} + a^{62}+.. + a + 1)  $
Read up on proof by induction and on geometric series.  They're fun stuff.
