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?
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 $$ p_{k+1}(x) = \sum\limits_{s=i}^{i+k}\prod\limits_{j=s+1}^{i+k}(x-\tau_j)[\tau_s,...,\tau_{i+k}]h.$$