#### Answer

$(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)(x-1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^{16}-1
,$ use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
x^{16}
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^{16}-1
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x^{8})^2-(1)^2
\\\\=
(x^{8}+1)(x^{8}-1)
\\\\=
(x^{8}+1)[(x^{4})^2-(1)^2]
\\\\=
(x^{8}+1)(x^{4}+1)(x^{4}-1)
\\\\=
(x^{8}+1)(x^{4}+1)[(x^{2})^2-(1)^2]
\\\\=
(x^{8}+1)(x^{4}+1)(x^{2}+1)(x^{2}-1)
\\\\=
(x^{8}+1)(x^{4}+1)(x^{2}+1)[(x)^{2}-(1)^2]
\\\\=
(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)(x-1)
.\end{array}