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?

You've basically got the right idea, except your last inequality is too weak (and not at all needed - you've basically just added an extra factor of $2$ without need). You've shown that
$$P_{k + 1} = P_k \cdot \frac{2k + 1}{k + 1}$$
Now we have that $P_k \le 2^k$ by hypothesis, and it's easy to see that
$$\frac{2k + 1}{k + 1} \le 2$$
for every $k \ge 0$. This implies $P_{k + 1} \le 2^k \cdot 2$, which is what you want.
