Can the Product of All Primes Before p Be Greater Than p^2?

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Homework Statement


I am proving something different and need this to be true.

choose prime p > 11. then p^2 is less than the product of all primes that came before it.


Homework Equations


U(n)= {1, a_1, ... a_k} this is the ring of numbers co prime to n.

ex: let p=13. 13^2 = 169<3*5*7*11

The Attempt at a Solution



I am using 11 because it's not generally true for primes less than 11 and I have dealt with those cases in my proof.

is this generally correct? is there a simple proof I should show? or take it as general knowledge.
 
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The "product of the first N primes" function grows so ridiculously fast as compared to the "square of the N-th prime" function, that pretty much any approximation at all should be usable to prove the inequality.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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