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

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!
