Lemma 43.15.4. Let $P$ be a polynomial of degree $r$ with leading coefficient $a$. Then

$r! a = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i{r \choose i} P(t - i)$

for any $t$.

Proof. Let us write $\Delta$ the operator which to a polynomial $P$ associates the polynomial $\Delta (P) = P(t) - P(t - 1)$. We claim that

$\Delta ^ r(P) = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i {r \choose i} P(t - i)$

This is true for $r = 0, 1$ by inspection. Assume it is true for $r$. Then we compute

\begin{align*} \Delta ^{r + 1}(P) & = \sum \nolimits _{i = 0, \ldots , r} (-1)^ i {r \choose i} \Delta (P)(t - i) \\ & = \sum \nolimits _{n = -r, \ldots , 0} (-1)^ i {r \choose i} (P(t - i) - P(t - i - 1)) \end{align*}

Thus the claim follows from the equality

${r + 1 \choose i} = {r \choose i} + {r \choose i - 1}$

The lemma follows from the fact that $\Delta (P)$ is of degree $r - 1$ with leading coefficient $ra$ if the degree of $P$ is $r$. $\square$

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