Is it easier to simplify it?

Relevant wiki: Multiplying and Dividing Monomials

$\dfrac{(x^2-3x+2)(x^2-4x+3)(x^2-5x+4)(x^2-6x+5)(x^2-7x+6)(x^2-8x+7)}{(x^2-x-2)(x^2-2x-3)(x^2-3x-4)(x^2-4x-5)(x^2-5x-6)(x^2-6x-7)}\\ =\dfrac{\color{#3D99F6}{\cancel{\color{#333333}{(x-2)}}}(x-1)\color{#D61F06}{\cancel{\color{#333333}{(x-3)}}}(x-1)\color{#EC7300}{\cancel{\color{#333333}{(x-4)}}}(x-1)\color{magenta}{\cancel{\color{#333333}{(x-5)}}}(x-1)\color{#20A900}{\cancel{\color{#333333}{(x-6)}}}(x-1)\color{#69047E}{\cancel{\color{#333333}{(x-7)}}}(x-1)}{\color{#3D99F6}{\cancel{\color{#333333}{(x-2)}}}(x+1)\color{#D61F06}{\cancel{\color{#333333}{(x-3)}}}(x+1)\color{#EC7300}{\cancel{\color{#333333}{(x-4)}}}(x+1)\color{magenta}{\cancel{\color{#333333}{(x-5)}}}(x+1)\color{#20A900}{\cancel{\color{#333333}{(x-6)}}}(x+1)\color{#69047E}{\cancel{\color{#333333}{(x-7)}}}(x+1)}\\ =\dfrac{(x-1)^6}{(x+1)^6}$

Substitute $x=-3$ into the expression:

$\dfrac{(-3-1)^6}{(-3+1)^6}\\ =\dfrac{(-4)^6}{(-2)^6}\\ =\dfrac{2^{12}}{2^6}\\ =2^6\\ =\boxed{64}$

To be less specific, let's consider these expressions : $N(x,n)=x^{ 2 }-(n+1)x+n$ and $D(x,n)=x^{ 2 }-(n-1)x-n$

We can easily compute that : $N(x,n)=(x-1)(x-n)$ and $D(x,n)=(x+1)(x-n)$

Therefore, $\frac { N(x,n) }{ D(x,n) } =\frac { x-1 }{ x+1 }$

The first expression can be rewritten as : $\prod _{ n=2 }^{ 7 }{ \frac { N(x,n) }{ D(x,n) } } ={ \left( \frac { x-1 }{ x+1 } \right) }^{ 6 }$

To answer the very first question, yes it is easier to simplify it :)

Substitute $x = -3$ then you get :

$\prod _{ n=2 }^{ 7 }{ \frac { N(-3,n) }{ D(-3,n) } } ={ \left( \frac { -4 }{ -2 } \right) }^{ 6 } = {2}^{6}=64$