TITLE: Does multiplying all a number's roots together give a product of infinity?
QUESTION [65 upvotes]: This is a recreational mathematics question that I thought up, and I can't see if the answer has been addressed either. 
Take a positive, real number greater than 1, and multiply all its roots together. The square root, multiplied by the cube root, multiplied by the fourth root, multiplied by the fifth root, and on until infinity. 
The reason it is a puzzle is that while multiplying a positive number by a number greater than 1 will always lead to a bigger number, the roots themselves are always getting smaller. My intuitive guess is that the product will go towards infinity, it could be that the product is finite. 
I also apologize that as a simple recreational mathematics enthusiast, I perhaps did not phrase this question perfectly.

REPLY [6 votes]: While the other answers are valid, they do tend to refer to previously established results, which (in a recreational context) is a bit frustrating. So here is a self-contained answer, for those who prefer that sort of thing.
First step: $$a^{1/2}\times a^{1/3}\times a^{1/4}\times…=a^{1/2+1/3+1/4+…}$$
Second step:$$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+…\text{equals:}$$
$$\frac{1}{2}\ge\frac{1}{2}$$
$$…+\frac{1}{3}+\frac{1}{4}\gt\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$…+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\gt\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
$$\text{…and so on.}$$
That is, $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+…$ is made up of infinitely many pieces, each of which is greater than $\frac12$. So it adds up to infinity, and so $$a^{1/2}\times a^{1/3}\times a^{1/4}\times…\text{ or}$$ $$a^{1/2+1/3+1/4+…}$$ is infinite if $a>1$, zero when $a<1$, and $1$ when $a=1$.
And not a $\sum$ sign anywhere to be seen!