Can the product of infinitely many elements from $\mathbb Q$ be irrational? I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. 
Thank you for your answers.
Yes, it can.

Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and
$$
  b_n = \frac{a_n}{a_{n-1}}
$$
for $n &gt; 1$.

We then have that
$$
  b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.
$$

We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.
There is a simple way to obtain any irrational number as an infinite product:


take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);
form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.


$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$