Q:

How does this simplification work?

The following recursive function was given:
$$T\left(n\right) = T\left(n - 1\right) + x$$
The author stated that by using repeated substitution we can solve the recurrence relation:

The basic idea is this: Given that $T\left(n\right) = T\left(n - 1\right) + x$, then we may also write $T\left(n - 1\right) = T\left(n - 2\right) + x$, provided $n>1$. Since $T(n-1)$ appears in the right-hand side of the former equation, we can substitute for it the entire right-hand side of the latter. By repeating this process we get:
\begin{align*}T\left(n\right) & = T\left(n - 1\right) + x \\
                                & = (T\left(n - 2\right) + x) + x \\
                                & = T\left(n - 2\right) + 2x \\
                                & = (T\left(n - 3\right) + x) + 2x \\
                                & = T\left(n - 3\right) + 3x
\end{align*}
The next step takes a little intuition: We must try to discern the pattern which is emerging. In this case it is obvious:
$$T\left(n\right) = T\left(n - k\right) + kx$$
where $1\leq k\leq n$.

So far so good. The author then wants to show an example in which we use an inductive process:

Inductive Hypothesis: Assume that $T\left(n\right) = T\left(n - k\right) + kx$ for $k = 1, 2,..., g$. By this assumption (call this $A$):
$$T\left(n\right) = T\left(n - g\right) + gx$$
Note also that using the original recurrence relation we can write (call this $B$):
$$T\left(n - g\right) = T\left(n - g - 1\right) + x$$

for $l \leq n$. Substituting Equation $A$ in the right-hand side of Equation $B$ gives:

\begin{align*}T\left(n\right) & = T\left(n - g - 1\right) + x + gx \\
                                & = T\left(n - \left(g + 1\right)\right) + \left(g + 1\right)x \\
\end{align*}

This is where he lost me. How did he go from $(n - g - 1)$ to $(n - (g + 1))$? 

A:

He didn't go from $g-1$ to $g+1$, he went from $-g-1$ to $-(g+1)$.  He simply factored out the $-1$ from both terms.
Recall that $-g$ denotes $(-1)\cdot g$, that $(-1)\cdot 1=-1$, and that $x-y$ is by definition $x+(-y)$.  By making the appropriate substitutes, we get that $-g-1$ is equal to $$(-1)\cdot g +(-1)=(-1)\cdot g+(-1)\cdot 1=(-1)\cdot (g+1)$$

