Negative factorials

$8! = 8×7×6×5×4×3×2×1\\ 6!= 6×5×4×3×2×1\\ 4! = 4×3×2×1$
So, we multiply till the first natural number $1$ .

But $-6$ itself is not a natural number. So, we cant multiply anything with it, hence it is not - defined.
So, $0! - \text{unfefined}$ = $1-\text{undefined} = \color{#69047E}{\boxed{\text{undefined}}}$ .

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Alternatively:-
We know that one factorial can be expressed in terms of another factorial of a greater number.
For example:- $7! = \dfrac{8!}{8}$ .

So, $(-6)!$ = $\dfrac{(-5)!}{(-5)}\\ = \dfrac{(-4)!}{-4 × -5}\\ = \dfrac{(-3)!}{-3 × -4 × -5}\\ = \dfrac{(-2)!}{-2 × -3 × -4 × -5}\\ = \dfrac{(-1)!}{-1 × -2 × -3 × -4 × -5}\\ = \dfrac{0!}{0 × -1 × -2 × -3 × -4 × -5}\\ = \dfrac{1}{0}\\ = \text{undefined}$

So, $0! - (-6)! = 1 - \text{undefined} = \color{#69047E}{\boxed{\text{undefined}}}$ .

The factorial function is not defined for negative integers. Thus the given expression is not defined.