If you define the value of an infinite operation as its limit, as mathematicians normally do, then you can make a limit argument for an infinite product of many numbers 0<n<1 being zero because each additional number that you multiply makes the product smaller, and you can find a point at which they will go smaller than any chosen positive number. In the epsilon-delta concept of limits (you challenge me to show where the value becomes closer than an epsilon of your choice to the purported limit; I show you where that happens) this then suffices to show that the limit is 0.

For example, suppose we multiply 0.9 by itself an infinite number of times. If we want it to be smaller than epsilon, we have to multiply it ceil(log_0.9(epsilon))+1 times. For instance, if you want it to get smaller than 0.01, the formula shows you should multiply it 45 times (actually 44 times suffices; the +1 is just for the case where it equals epsilon exactly). Here 0.9⁴⁵=0.00872796356808771242 which is indeed smaller than 0.01.

Correct. For example the series (1/n)(1/n+1)(1/n+2)... converges to zero as n-> ∞ (actually the term is "diverges to zero" for products)

You can play around with different series on Wolfram Alpha https://www.wolframalpha.com/input/?i=infinite+product+1%2Fn

Consider the product of the sequence which has ½ repeated N times. The product is (½)^N; which halves each time. If we want the product to go below x, then we can plug in N > log2(1/x). Since for any x > 0 we can find an N, we say lim (½)^∞ = 0.

tl;dr Basically, as N goes to infinity, the product gets arbitrarily close to 0.

Yes. Easiest example would be just the product of a_n, n=1,2,3,..., with a_n=same constant in (0,1). Then all the partial products will be positive, but the limit is clearly 0.

Yes -- just look at Pi_0^inf(.1), the partial products are just 0.1^n, and the limit of that as n goes to infinity is zero.