Summation, capital pi properties I know that for summation,
$$\sum_{i = 1}^{x} a_{i} + \sum_{i = x+1}^{n} a_{i} = \sum_{i = 1}^{n} a_{i}$$ 
Does the same concept apply for capital pi? Such as:
$$\prod_{i = 1}^{x} a_{i} \cdot\prod_{i = x+1}^{n} a_{i} = \prod_{i = 1}^{n} a_{i}$$ 
I am trying to show that 
$$\prod_{i = 1}^{0} a_{i} = 1$$

The notation $\displaystyle\prod_{i=n}^m a_i$ can also be written as $\displaystyle\prod_{i\in\{n,..,m\}}a_i$.  
This notation can be generalised as $\displaystyle \prod_{i\in A}a_i$ for any set of indices, $A\subseteq\Bbb N^+$.
We do want the following to hold for any disjoint sets of indices, $A, B$.
$$\prod_{i\in A\cup B} a_i = (\prod_{i\in A} a_i)\cdot(\prod_{i\in B} a_i)$$
Since the empty set is disjoint with any set, then this means we must have: $$\prod_{i\in A\cup\emptyset} a_i = (\prod_{i\in A} a_i)\cdot(\prod_{i\in \emptyset} a_i)$$
Because $A\cup\emptyset = A$, then clearly the empty product needs to equal $1$. $$\prod_{i\in\{\}} a_i = 1$$

PS: by the same reasoning, the empty series equals $0$. $$\sum_{i\in\{\}} a_i = 0$$
