Convolving time reversed discrete input with time reversed discrete impulse response
Here is another example of how incredibly poor choice of notation misleads students everywhere. I do not possess a copy of any of the various tomes on signal processing that seem to be revered as the fifth Gospel on this site, and so
I do not know if the book cited by the OP actually has written $(1)$,
but I assert that it is against mathematical common sense to write things like
$$y[n] = x[n] \star h[n].\tag{1}$$
It is generally accepted mathematical convention that is a symbol
has the same meaning everywhere it appears in an equation or expression. 
For example, when we write
$$y(t) = x(t)h(t),~ -\infty &lt; t &lt; \infty, \tag{2}$$
we mean by this that for every choice of real number $t$, the value
of $y(t)$ is the same as the product of the values of $x(t)$ and $h(t)$;
for example, $y(3) = x(3)h(3)$, and $y(313.012) = x(313.012)h(313.012)$,
ans so on and so forth.
So how are we supposed to make sense of this monstrosity $(1)$?
Clearly $y[3]$ is not $x[3]\star h[3]$ whatever meaning we might
ascribe to the latter quantity; the value of $y[3]$ depends on
a lot of other values of $x$ and $h$.  The poor notation misleads
students into writing things like

$$\sum_{k=-\infty}^{\infty}{\left((-1)^{k}(u[-k]-u[-k-8])\right)\left(u[-n-k]-u[(-n-8)-k]\right)}$$

which is not the correct expression for the convolution (or the correlation,
for that matter) of 
$$x[n]=(-1)^n{u[-n]-u[-n-8]}\quad \text{and} \quad
h[n]=u[n]-u[n-8]\tag{3}$$

The correct way to do this is write
$$y[n] = (x\star h)[n] = \sum_{k=-\infty}^\infty x[k]h[n-k]$$
and then substitute $n-k$ for $n$ in $h[n]=u[n]-u[n-8]$ 
to
get $h[n-k] = u[n-k]- u[n-k-8]$ and not $u[-n-k]-u[(-n-8)-k]$
the way that the OP has it.  

Some initial thought might have
shown that (assuming that $u[n]$ is the unit step),
that $h[n]$ is $0$ for $n &lt; 0$ (since then both $u[n]$ and $u[n-8]$
are $0$, and that $h[n] = 0$ for $n \geq 8$ (since then
both $u[n]$ and $u[n-8]$ are $1$. Consequently for any given 
fixed integer $n$, $h[n-k]$ is nonzero only for
$$0 \leq n-k \leq 7 \Rightarrow n-7 \leq k \leq n$$
and so the limits on the sum could be reduced from
$-\infty$ and $\infty$ to $n-7$ and $n$ (cf. my comment
on the OP's question).  I will leave it to the OP to figure
out whether the fact that $x[n]$ is also time-limited
can be used to further reduce the limits on the sum.
Hint: the answer to this might depend on the specific
value of $n$, and you may need to try several different
values of $n$, e.g. $0, 3, 8, 10, -40$ etc to see if you
can discern a pattern.