A morphism of coalgebras related to Binomial coefficients I am looking for a hint to prove that
$$\Delta({t\choose n})= \sum\limits_{i=0}^{n} {t\choose i} \otimes {t\choose n-i} $$
where ${t\choose k}= \frac{t(t-1)...(t-k+1)}{k!}$ is a polynomial in $t$, $\Delta$ being the comultiplication in the coalgebra of polynomials $k[t]$ .
This is equivalent to show that $\quad \frac{1}{n!} \prod\limits_{j=0}^{n- 1} \Delta(t-j)= \sum\limits_{i=0}^{n} {t\choose i} \otimes {t\choose n-i} \\ 
\text { or }\quad \ \prod\limits_{j=0}^{n- 1} \Delta(t-j)= n! \sum\limits_{i=0}^{n} \frac{t(t- 1)... (t- i+ 1)}{i!} \otimes \frac{t(t- 1)... (t- n+i + 1)}{(n-i)!} \\
  \quad\quad\quad\quad\quad\ \quad\quad\quad  = \sum\limits_{i=0}^{n} {n \choose i}\ t(t- 1)... (t- i+ 1) \otimes t(t- 1)... (t- n+ i+ 1)$
Induction doesn't seem to be the way to go, is there an argument where we should say that $\Delta({t\choose n})= \Delta(\frac{t(t-1)...(t-n+1)}{n!})= \Delta(\frac{t(t-1)...(t-i+1)}{i!})\Delta(\frac{(t- i)(t-i-1)...(t-n+1)}{n(n-1)...(i+1)})$ splits somehow into the desired sum?
 A: Here is a hint, which avoids most of the nasty computation that this problem could otherwise entail:
First, show that the falling factorials
$$t^{\underline{n}} \triangleq t (t-1) (t-2) \ldots (t-(n-1))$$
form a Sequence of Binomial Type, in the sense that
$$(x+y)^{\underline{n}} = \sum_k \binom{n}{k} x^{\underline{k}} y^{\underline{n-k}}$$
(officially we need to check that $\text{deg}(t^\underline{n}) = n$ too, but this is clear). There's a slew of ways you can check this, see here for instance.
Next, since $\Delta$ is an algebra hom, notice
$$
\Delta (t^\underline{n}) = 
\left ( \Delta t \right )^\underline{n} = 
\left ( t \otimes 1 + 1 \otimes t \right )^\underline{n}
$$
and from here, a routine calculation will give you what you want.
As an aside, it shouldn't be surprising that the core of this argument is combinatorial. The comultiplication of hopf algebras is extremely useful for modelling the ways you can "subdivide" a combinatorial object, and this problem is one manifestation of that fact. See, for instance, Joni and Rota's Coalgebras and Bialgebras in Combinatorics for more.

I hope this helps ^_^
