Factorial solve without substitution [closed]

I'm in stuck with this dimostration. I've got $$\frac{n!}{(n+1)!}$$ and it's must be $$\frac{1}{n+1}$$ If I put n=3, I've got $$\frac{3!}{(3+1)!}=\frac{1}{4}=\frac{1}{3+1}$$ and it's correct. But I can't prove the resolution without replacing, like in this case that I put n=3. Can someone explain to me how to do it?

closed as unclear what you're asking by José Carlos Santos, Cesareo, Hayk, Xander Henderson, metamorphyMay 22 at 16:07

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• Please fix the typesetting. – Yves Daoust May 21 at 10:06
• How do you define the factorial ? – Yves Daoust May 21 at 10:07
• What Yves means, is that you can use the definition of factorial (plug it in the expression) and you should be able to proceed. – Matti P. May 21 at 10:09
• The definition is n! =n(n-1)! And n! /(n-k)! =n(n-1)(n-k+1) – Ciao May 21 at 10:15
• Literally the first thing you wrote, yields the result. – maxmilgram May 21 at 10:18

Recall the definition of factorial, $$n! = 1 \times 2 \times \ldots \times n$$. So we have $$(n+1)! = 1 \times 2 \times \ldots \times n \times (n+1) = n! \times (n+1)$$. But from $$n! \times (n+1) = (n+1)!$$ we simply rearrange to get the desired result.