I read the following statement:
We can define the number $$x=2^0\cdot 3^1\cdot 5^2\cdot 7^3\cdot\ldots\cdot b^n$$ where $b$ is the $n$'th prime number. That is, $b$ is the $n$'th prime number if and only if there is some $x<b^{2n}$ that can be factored as above and has $b^n$ but not $b^{n+1}$ as a factor.
It's not immediately obvious to me how $x<b^{2n}$ (but maybe there's something I'm not seeing properly). How would one prove this inequality?
EDIT: It has come to my attention that the initial statement was $x<{b^n}^2$, which was meant to be read as $x<b^{(n^2)}$ and not as $x<\left(b^n\right)^2$ (the way I used in this question).
First, I'd like to apologize for making a mistake in the question - then again, the original formulation in the text was ambiguous. The edited property is now easy to prove.
For $n>1$, we have that $$\sum_{i=1}^{i=n}i=\frac{n(n+1)}{2}<n^2.$$ Therefore: $$x=2^0\cdot 3^1\cdot 5^2\cdot\ldots\cdot b^n<b^0\cdot b^1\cdot b^2\cdot\ldots \cdot b^{n-1}\cdot b^n=b^{\sum_{i=1}^{i=n} i}<b^{(n^2)}$$