A Practical Guide to Splines (De Boor) - Proof of Leibniz formula In De Boor's A Practical Guide to Splines (1978) Leibniz' formula is defined as follows (p.5):

If $f = gh$, i.e. $f(x) = g(x)h(x)$ for all x, then
  $$ 
[\tau_i, ..., \tau_{i+k}]f = 
\sum\limits_{r = i}^{i+k}([\tau_i,...,\tau_r]g)([\tau_r,...,\tau_{i+k}]h).
$$

Where $[\tau_i,...,\tau_{i+k}]f$ denotes the k-th divided difference of $f$ at the points $[\tau_i,...,\tau_{i+k}]$ which is the leading coefficient of the polynomial $p$ such that $(\forall j \in [i, i+k]) \: p(\tau_j) = f(\tau_j)$.
The proof starts with pointing out that the newly introduced function

$$
F(x) = \sum\limits_{r=i}^{i+k} \prod_{j=i}^{r-1}(x - \tau_j)[\tau_i,...,\tau_r]g
\sum\limits_{s=i}^{i+k} \prod_{j=s+1}^{i+k}(x - \tau_j)[\tau_s,...,\tau_{i+k}]h
$$
  agrees with $f$ at $\tau_i, ..., \tau_{i+k}$ since, by (4), the first factor agrees with $g$ and the second factor agrees with $h$ there.

Where (4) is the Newton form of a polynomial $p_n$ of order $n$:

$$
p_n(x) = \sum\limits_{i=1}^n \prod\limits_{j=1}^{i-1}(x-\tau_j)[\tau_1,...,\tau_i]g,
$$ 

that agrees with $g$ at the points $\tau_1, ..., \tau_n$.
My problem is that I don't see why $F$ should agree with $f$ at $\tau_i,...,\tau_{i+k}$. 
If, e.g., I choose $k=1$, I get
$$ F(x) = \sum\limits_{r=i}^{i+1} \prod_{j=i}^{r-1}(x - \tau_j)[\tau_i,...,\tau_r]g
\sum\limits_{s=i}^{i+1} \prod_{j=s+1}^{i+1}(x - \tau_j)[\tau_s,...,\tau_{i+1}]h, $$ or, equivalently,
$$ F(x) = ([\tau_i]g + (x - \tau_i)[\tau_i, \tau_{i+1}]g)((x - \tau_{i+1})[\tau_i, \tau_{i+1}]h + [\tau_{i+1}]h).$$
Evaluating for $x = \tau_i$ yields
$$ F(\tau_i) = g(\tau_i)((\tau_i - \tau_{i+1})[\tau_i, \tau_{i+1}]h + h(\tau_{i+1}))
\neq f(\tau_i) = g(\tau_i)h(\tau_i). $$
What went wrong?
 A: What went wrong is that the last inequality is actually an equation:
$$ 
\begin{align}
F(\tau_i) & = g(\tau_i)((\tau_i - \tau_{i+1})[\tau_i, \tau_{i+1}]h + h(\tau_{i+1})) \\
& = g(\tau_i)
\left(
\frac{\tau_i - \tau_{i+1}}{\tau_{i+1} - \tau_i}
(-h(\tau_i)+h(\tau_{i+1}))
+h(\tau_{i+1})
\right) \\
& = g(\tau_i)h(\tau_i) \\
& = f(\tau_i).
\end{align}$$
Note:
The reason why the $2^{\text{nd}}$ factor in $F(x)$ agrees with $h$ at $\tau_i,...,\tau_{i+k}$, though defined differently than the Newton form given in (4), is that generally the Newton form of a polynomial is described via
$$ p_{k+1}(x) = p_{k+1}(x) + 
(p_k(x) - p_k(x))
 + ... + 
(p_1(x) - p_1(x))
,
$$
such that the points $\tau_{j_\xi}$ with $(\forall m \in [1,k])(\forall \xi \in [1,m]) \: p_m(\tau_{j_\xi}) = h(\tau_{j_\xi})$ can be arbitrarily chosen. 
Given the following choice of points where $p_j$ agrees with $h$:
$$ 
\begin{align}
& p_1(\tau_{i+k}) = h(\tau_{i+k}) \\
& (p_2(\tau_{i+k}) = h(\tau_{i+k})) \wedge (p_2(\tau_{i+k-1}) = h(\tau_{i+k-1})) \\
\vdots \\
& (p_{k+1}(\tau_{i+k}) = h(\tau_{i+k})) \: \wedge \: ... \: \wedge \: (p_{k+1}(\tau_i) = h(\tau_i)),
\end{align}
$$
property (i) on p.3 requires that
$$ p_{k+1}(x) = \sum\limits_{s=i}^{i+k}\prod\limits_{j=s+1}^{i+k}(x-\tau_j)[\tau_s,...,\tau_{i+k}]h.$$
