# $\prod$ and factorial

$$\prod_{i=0}^{j-1}(j-i+1)$$ is $$=$$ to $$(j+1)!$$ or $$\le$$? I think it is $$=$$, but if it's not, please put an explanation.

• Yes, it is equal to that (case $j=0$ needs special attention, though). – Saucy O'Path May 11 at 16:55
• Expand the product what do you get? – kingW3 May 11 at 16:57

$$\prod_{i=0}^{j-1}(j-i+1)=(j+1)\cdot j\cdot (j-1)\cdots 3\cdot 2=(j+1)!$$
\begin{align} \prod_{i=0}^{j-1} (j-i+1) & = (j+1)\cdot j\cdot (j-1)\cdots(j-(j-2)+1)\cdot (j-(j-1)+1)\\ & = 2\cdots (j+1)=(j+1)! \end{align}
Set $$j-i+1=t$$
$$i=0\implies t=j+1$$
$$i=j-1\implies t=?$$
$$\prod_{i=0}^{j-1}(j-i+1)=\prod_{t=j+1}^2t=\prod_{t=2}^{j+1}t=?$$