How to show that $(n-k)!n^{\underline{k}}=\Gamma(n+1)$

Question

I want to show that, $\forall n,k\in\mathbb{N}$: $${{n}\choose{k}}=\dfrac{n^{\underline{k}}}{k!}$$ And I have the following with $\Gamma(n)=(n-1)!$: \begin{align*} {{n}\choose{k}}-\dfrac{n^{\underline{k}}}{k!}&=\dfrac{n!}{k!(n-k)!}-\dfrac{n^{\underline{k}}}{k!}\\ &=\dfrac{n!}{k!(n-k)!}-\dfrac{(n-k)!n^{\underline{k}}}{k!(n-k)!}\\ &=\dfrac{\Gamma(n+1)}{k!(n-k)!}-\dfrac{\Gamma(n+1)}{k!(n-k)!}, \text{ thanks to WolframAlpha}\\ &=0 \end{align*}

But how did WolframAlpha to show that $(n-k)!n^{\underline{k}}=\Gamma(n+1)$??

Answer 1

Note

$$n^{\underline{k}}=n \times (n-1) \times (n-2) \times \dots (n-k+1) \tag{1}$$ Also $$k!=k \times (k-1) \times (k-2) \times \dots \times 1 \tag{2}$$ So since $$n!=\{n \times (n-1) \times (n-2) \times \dots (n-k+1)\} \times \{k \times (k-1) \times \dots\times 1\}$$ This is just $(1) \times (2)$.