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

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  • $\begingroup$ Yes, it is equal to that (case $j=0$ needs special attention, though). $\endgroup$ – Saucy O'Path May 11 at 16:55
  • $\begingroup$ Expand the product what do you get? $\endgroup$ – kingW3 May 11 at 16:57
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$$ \prod_{i=0}^{j-1}(j-i+1)=(j+1)\cdot j\cdot (j-1)\cdots 3\cdot 2=(j+1)! $$

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\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}

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Set $j-i+1=t$

$i=0\implies t=j+1$

$i=j-1\implies t=?$

$$\prod_{i=0}^{j-1}(j-i+1)=\prod_{t=j+1}^2t=\prod_{t=2}^{j+1}t=?$$

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