<s>[INST] Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n  $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument: 
$P_{k+1} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)\cdot(2k+1)}{1\cdot2\cdot3\cdot\ldots\cdot k \cdot (k+1)} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)}{1\cdot2\cdot3\cdot\ldots\cdot k}\cdot \frac{2k+1}{k+1} \leq 2^k \cdot\frac{2k+1}{k+1} \leq 2^{k+1}\cdot \frac{2k+1}{k+1}$ using $P_k$ and that these are all positive numbers.
Now dividing through by $\frac{2k+1}{k+1}$ would give the desired result.
I just feel like I've done something wrong here. Can anyone tell me if I've overlooked anything?
 [/INST] Here is an alternate solution that gives a bit more insight.
$\frac{1 \cdot 3 \cdots (2n-1)}{n!} = \frac{(2n)!}{2^n(n!)^2}$ so we have to prove $2^{2n} \ge \frac{(2n)!}{(n!)^2} = \dbinom{2n}{n}$.
However, $2^{2n} = \sum_{k = 0}^{2n} \dbinom{2n}k > \dbinom{2n}n$ and the conclusion follows. Obviously, the inequality is very weak.  
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