<s>[INST] Show that $\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$ If $n$ is a positive integer, show that

$\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$

This is how I tried to solve it:
the given inequality is equivalent to 
$\binom{n}{1}...\binom{n}{n-1}\leq ({\frac{2^n-2}{n-1}})^{n-1}$
$\iff \sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}}\leq \frac{2^n-2}{n-1}$
Now, we know that 
$\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{\binom{n}{1}+...+\binom{n}{n-1}}{n-1}$
Hence, we are left with proving $\binom{n}{1}+...+\binom{n}{n-1} \leq 2^n-2$
or, $\binom{n}{0}+...+\binom{n}{n} \leq 2^n$
Can we prove the last inequality? Am I on the right track? Thanks for help.
 [/INST] As noted by @LeGrandDODOM from AM-GM that $\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{\binom{n}{1}+...+\binom{n}{n-1}}{n-1}$.
However, note that $\binom{n}{1}+...+\binom{n}{n-1}=2^n-2$ which follows from the binomial theorem. 
$\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{2^n-2}{n-1}$. 
This gives us that ${\binom{n}{1}...\binom{n}{n-1}} \leq (\frac{2^n-2}{n-1})^{n-1}$.
From here, $\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$.
Thus, you appear to mistaken. 
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