Why is $\prod_{n}^{n-1} = 1$ and $\sum_{k=j+1}^{j}=0$? 
The general first-order difference equation has the form
  $$
x_{n+1}= a_n x_n + g_n, \quad n\geq 0, \tag 1
$$
  where $(a_n)_{n\in \mathbb N_0}$ and $(g_n)_{n\in \mathbb N_0}$ are given sequences.
  The solution is given by
  $$
x_n = x_0 \prod_{k=0}^{n-1} a_k + \sum_{k=0}^{n-1}g_k \prod_{i=k+1}^{n-1}a_i
$$
  where we used
  \begin{align}
\prod_{n}^{n-1} = 1 \tag 2 \\
\sum_{k=j+1}^{j}=0 \tag 3
\end{align}

I don't follow $(2)$ and $(3)$, isn't
\begin{align}
\prod_{n}^{n-1}1 = n \cdot (n-1) \quad ?\tag 4
\end{align}
And 
\begin{align}
\sum_{k=j+1}^{j}1= j \quad ?\tag 5
\end{align}
 A: There are inconsistencies in your question statement.
In the first place, the author does specify that in case of reversed bounds you get a $0$ sum or a $1$ product. If he gives you his conventions $(2)/(3)$, you should not question them. 
Then if you reject the convention and consider that reversed ranges are not equivalent to empty, you should write
$$\prod_{\color{green}{k=}n}^{n-1}1 = 1\cdot1=1 \quad \tag 4$$
and 
$$\sum_{k=j+1}^{j}1= 1+1=2\tag 5.$$
[In green, my own addition.]
A: If $$\sum_{k=m}^{n+1}a_k = a_{n+1}+\sum_{k=m}^{n}a_k$$ and for example $$\sum_{k=m}^{m+2}a_k=a_m+a_{m+1}+a_{m+2},$$ then we have no "choice" but acknowledging that
$$ \sum_{k=m}^{m+1}a_k=\sum_{k=m}^{m+2}a_k\;-a_{m+2}=a_m+a_{m+1},$$
$$ \sum_{k=m}^{m}a_k=\sum_{k=m}^{m+1}a_k\;-a_{m+1}=a_m,$$
$$ \sum_{k=m}^{m-1}a_k=\sum_{k=m}^{m}a_k\;-a_{m}=0.$$
We might even go beyond this and define sums where the top index is way less than the lower index, But I won't delve into that.
In the end, it is a matter of definition.
Given a set $S$ and a map $f\colon S\to A$ into an abelian group $A$, how can we define something that is worth being called the sum of function values of $f$, and for which we want to employ the notation
$$ \sum_{x\in S}f(x)?$$
We do it by recursion and the start is the case of the empty set! We define
$$ \sum_{x\in \emptyset}f(x)=0$$
because after all $0$ is the only element of $A$ we can put our finger on.
If $S$ is non-empty, say there is some $s\in S$, then $S\setminus\{s\}$ is smaller and we may try to use recursion to define
$$ \sum_{x\in S}f(x)=f(s)+\sum_{x\in S\setminus\{s\}}f(x).$$
Fortunately (or thanks to commutativity and associativity of addition in $A$), this is well-defined, i.e., does not depend on the choice of $s$. On the other hand, this recursion works only for finite sets $S$. 
But at least, this case is perhaps less confusing that the common two-index notation $\sum_{i=n}^m$ for $\sum_{i\in \{\,k\in\Bbb Z\mid n\le k\le m\,\}}$.
