Why is $\prod_{i=1}^p(np+i)=(np+p)!/(np)!$, where $n\geq 0,p\geq 1$ are integers?

Question

I checked on WolframAlpha whether or not if $(np+p)!=(np)!\prod_{i=1}^p(np+i)$ is true, but it indeed does. I spent a lot of time to verify it, but I could not do it because of what $(x+y)!$ would mean for $x,y\in \mathbb{N}$. However, in the end, I managed to figure it out by rewriting $j!$ into $\exp(\sum_{j=1}^n\ln(j))$ for $j\geq 1$, and then proceed it. The above formula is clear for $n=0$, so we may consider $n\geq 1$. Since we can write $$ \sum_{j=1}^{np+p}\ln(j)=\sum_{j=1}^{np}\ln(j)+\sum_{j=np+1}^{np+p}\ln(j)=\sum_{j=1}^{np}\ln(j)+\sum_{m=1}^{p}\ln(np+m), $$ we conclude that $$ (np+p)!=\exp\Big(\sum_{j=1}^{np+p}\ln(j)\Big)=\exp\Big(\sum_{j=1}^{np}\ln(j)\Big)\exp\Big(\sum_{m=1}^{p}\ln(np+m)\Big)=(np)!\prod_{m=1}^p(np+m). $$ Is this the correct way of proving? If not, how would you prove it?

Answer 1

$$(np+p)!=1\cdot 2\cdot...\cdot(np+p)=1\cdot 2\cdot ... \cdot(np)\cdot (np+1)\cdot ... \cdot (np+p)\\=(np)!\cdot \prod_{m=1}^p (np+m),$$ So $$\prod_{m=1}^p(np+m)=(np+p)!/(np)!.$$