Factorial everywhere

$\begin{aligned} X & = \frac {1!\times 2! \times 3! \cdots \times 10!}{(1!)^2(3!)^2(5!)^2(7!)^2(9!)^2} \\ & = \frac {2!\times 4! \times 6! \times 8! \times 10!}{3! \times 5! \times 7! \times 9!} \\ & = 2 \times 4 \times 6 \times 8 \times 10 \\ & = 2^{1+2+1+3+1} \times 3 \times 5 \\ & = 15 \times 2^8 \end{aligned}$

$\implies n = \boxed{8}$

$\frac { 1! × 2! × 3! × \ldots × 9! × 10! } { (1!)^2 × (3!)^2 × (5!)^2 × (7!)^2 × (9!)^2 } = 15 × 2^n$

$\frac { 1! } { 1! } × \frac { 2 × 1! } { 1! } × \frac { 3! } { 3! } × \frac { 4 × 3! } { 3! } × \frac { 5! } { 5! } × \frac { 6 × 5! } { 5! } × \frac { 7! } { 7! } × \frac { 8 × 7! } { 7! } × \frac { 9! } { 9! } × \frac { 10 × 9! } { 9! } = 15 × 2^n$

After simplifying:

$2 × 4 × 6 × 8 × 10 = 15 × 2^n$

$2 × 2^2 × (2 × 3) × 2^3 × (2×5) = 15 × 2^n$

$15 × 2^8 = 15 × 2^n$

$n = \boxed {8}$