Does multiplying all a number's roots together give a product of infinity?

This is an interesting question, but thanks to the properties of exponentiation, it can be easily solved as follows: $$\prod_{k\ge1}\sqrt[k]{n}=\prod_{k\ge1}n^{1/k}=n^{\sum_{k\ge1}\frac{1}{k}},$$ and as the exponent diverges (Harmonic series) the whole expression diverges, as long as $n>1$.

EDIT: the first equality is definition. For the second equality we need to argue why one can pass to the limit.

In other words: for any fixed $N$ the equality $$\prod_{k=1}^N n^{1/k}=n^{\sum_{k=1}^N \frac{1}{k}},$$ is valid, but why does it hold that $$\lim_{N\rightarrow \infty}n^{\sum_{k=1}^N \frac{1}{k}}=n^{\lim_{N\rightarrow\infty}\sum_{k=1}^N \frac{1}{k}}\quad ?$$ In this particular case it is a consequence of the fact that the map $x\mapsto n^x$ diverges to $+\infty$ as $x\rightarrow +\infty$ and by abuse of notation we write $n^{+\infty}=+\infty$

If conversely the exponent would converge to a finite limit, as for example in the case mentioned in the comment of IanF1, then this would be a simple consequence of the fact that the map $x\mapsto n^x$ is continuous.

What you are describing can be written as $$\prod_{n=1}^\infty{x^{1/n}}=x^1\times x^{1/2}\times x^{1/3} \times x^{1/4}...$$ Because to multiply a number to several powers is equivalent to adding their powers, this can be rewritten as $$x^{y}$$ where $$y={\sum_{n=1}^\infty{1/n}}$$

If we only added up the first $m$ terms of this, we get what's called the $m$th harmonic number. The harmonic numbers go 1, 3/2, 11/6, 25/12, ... Each one is about as big as the natural logarithm of $m$.

However since we're adding infinitely we are heading towards the "last" or "highest" harmonic number. Sadly, there is no greatest harmonic number and there is no point where they get ever-closer to some finite number so we are in some sense heading towards $x^\infty$.

We say that a series like this, i.e. which does not converge on some finite number "does not converge". And this makes it very difficult to quantify its size in a meaningful way.

However if however you want to get an idea of how quickly it diverges, you might imagine that the product of the first $m$ numbers is approximately equal to:

$$x^{\log_e{m}}$$

So if you had some number $p$ that measured the number of "infinite" terms you are multiplying together, then you might imagine your number is in some very limited sense $$x^{\log{p}}$$

From this you can observe that if you were to allow $x=e$ then your series equals $p$, so in a sense it reaches infinity at the same rate as you add terms. For $x=2$ it lags $m$ in approaching infinity, and for $x\geq3$ it is ahead of $m$ in approaching infinity.

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!