Factorial canceling on expansion of binomial coefficients on Concrete Mathematics On Concrete Mathematics section 5.5, which is teaching the hypergeometric functions, generalized factorials is defined as:
\[
\frac 1 {z!} = \lim_{n \to \infty} \binom{n+z}{n}n^{-z}
\]
where
\[
\binom r k =
\left\{
\begin{array}{ll}
r^{\underline k} / k! = r(r-1) \cdots (r-k+1) / k! & k > 0 \\
1 & k = 0 \\
0 & k < 0
\end{array}
\right.
\]
follows the ordinary definition.
So $z! = z(z-1)!$ for all complex $z$ (except negative integers), then we can check $0! = 1$ and $n! = n(n-1) \cdots 1$ for $n > 0$.
Then, a binomial coefficient can be written
\[
\binom z w = \lim_{\zeta \to z} \lim_{\omega \to w} \frac{\zeta!}{\omega!(\zeta-\omega)!}
\]
Let $t_k = \dbinom r k \dbinom s {n-k}$.
However, the succeeding paragraph says that

\[ t_k = \frac{r!}{(r-k)!k!} \frac{s!}{(s-n+k)!(n-k)!}\]
  and we are no longer too shy to use generalized factorials in these expressions.

without limits (it is said that we must use appropriate limiting values when these formulas give $\infty / \infty$) and considers the ratio $t_{k+1} / t_k$ for all $t_k \neq 0$ and cancels some factorials using the property $z! = z(z-1)!$
I'm "too shy" and my question remains: why can we do such canceling?
To observe closely, we take a variety of an example from section 5.7:
Considering indefinite summation
 \[
 \sum \binom n {-k} \delta k, \qquad n < 0
 \]
Let $t(k) = \dbinom n {-k} = \dfrac{n!}{(-k)!(n+k)!}$, we have
 \[
 \frac{t(k+1)}{t(k)} = \frac{n!}{(-k-1)!(n+k+1)!} \frac{(-k)!(n+k)!}{n!} = -\frac{k}{n+k+1}
 \]
Let $n = -1$, we have $t(k+1) / t(k) = -1$ for $t(k) \neq 0$. But it's wrong for $k = 0$, where $t(1) = 0$ and $t(0) = 1$.
To see how the error happens, we resume the $\lim$ notation:
\begin{align*}
 t(k+1)
  &= \binom n {-k-1} \\
  &= \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{(n+z_2)!}{(-k-1+z_1)!(n+k+1-z_1+z_2)!} \\
  &= \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{-k+z_1}{n+k+1-z_1+z_2} \frac{(n+z_2)!}{(-k+z_1)!(n+k-z_1+z_2)!} \\
  &= \binom n {-k} \lim_{z_2 \to 0} \lim_{z_1 \to 0} \frac{-k+z_1}{n+k+1-z_1+z_2} 
 \end{align*}
 So when $n = -1$ and $k = 0$, we have $\lim_{z_2 \to 0} \lim_{z_1 \to 0} (-k+z_1)/(n+k+1-z_1+z_2) = 0$ not $-k/(n+k+1)$.
Another example (also from section 5.5) is:
\[
  \lim_{x \to -1} \frac{x!}{(x-1)!} = \lim_{x \to -1} x = -1
 \]
 but
 \[
  \lim_{x \to -1} \frac{x!}{(2x)!} = -2
 \]
 because of $(-z)! \Gamma(z) = \pi / \sin(z\pi)$, so expression $(-2)! / (-1)!$ is illegal.
My question is: in the frame of Concrete Mathematics, how to prevent such errors?
Thanks a lot.
 A: If the factorials in the limit expression differ by an integer, then cancel away.
But if the difference only becomes an integer in the limit, then you obviously cannot cancel before taking the limit, so you are not allowed to cancel at all.
The difference between $x$ and $2x$ is $x$ which is not an integer until the limit of $-1$ is reached.
A: I had very similar doubts but was unable to express them as well as you did.
If I understand this correctly, to know when cancellation of factorials is valid, it suffices to consider the validity of the equality $\binom{n}{k}=\frac{n!}{k!(n-k)!}$---(1). This is because the recursive relation $x!=x(x-1)!,x\notin \mathbb{Z}^-$ is established before the expression$\frac{n!}{k!(n-k)!}$ is stated. So any algebraic manipulation between $\binom{n1}{k1},\binom{n2}{k2}$ is the same as that between $\frac{n1!}{k1!(n1-k1)!}\frac{n2!}{k2!(n2-k2)!}$ as long as (1) holds for $\binom{n1}{k1},\binom{n2}{k2}$.
Assume $k\in \mathbb{Z}$, in my opinion this is a valid assumption as hypergeometric functions are a sum over integers $k\geq 0$.
I will consider $n\in \mathbb{R}$, not quite sure if it extends to $\mathbb{C}$ as well.
Case 1. If $n\in \mathbb{R}\setminus \mathbb{Z}$, then $(n-k)\in \mathbb{R}\setminus \mathbb{Z}$, hence the equality (1) holds since $n!,(n-k)!\in (-\infty, \infty)$.
Case 2 $n\in \mathbb{Z}$
Equality (1) only breaks when we have $|n!|=+\infty$ and $|k!| or |(n-k)!|=+\infty$ at the same time $\iff n<0$ and $n<k$ or $k\leq n$, i.e. $n<0$. 
An example when (1) holds is where $n\in \mathbb{Z}^+_0, n-k<0$, then clearly $n^{\underline{k}}=0=n!\frac{1}{(n-k)!}$.
Hence direct cancellation is invalid $\iff$ $n\in \mathbb{Z}$, $n<0$ for any binomial coefficient $\binom{n}{k}$ involved.
