Prove $\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$ without induction

Question

I found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.

Prove for odd $n$ that $$\sum_{k=1}^{(n+1)/2}\prod_{j=0}^{k-1} \left(\frac{n-2j-1}{n-2j}\right)=\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$

Answer 1

We need to prove that $$\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$$ or $$1+\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n}{3}$$ or $$\frac{n-3}{n-2}+ \frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-3}{3}$$ or $$1+\frac{n-5}{n-4} +\frac{n-5}{n-4}\frac{n-7}{n-6} ... = \frac{n-2}{3}$$ or $$.$$ $$.$$ $$.$$ $$1+\frac{2}{3}=\frac{5}{3}.$$ Done!