I'll assume . For it can be checked directly.
Let
Then
Differentiating more, it's easily seen that
Where on the left side we have differentiations and on the right side are polynomials of degree .
As such, form a basis to the vector space of polynomials of degree at most . Therefore,
is a linear combination of
up to times differentiation.
But since the zero of at is of order , we have that all the functions in (1) evaluated at vanish, and thus their linear combination evaluated at that point,
also does.
My solutions:
1.
First proof: Consider the set of functions from a set .
If we fix a subset we have that the number of functions from such that is so in fact what we are counting (in the original equation) by inclusion-exclusion is the number of functions such that which is of course absurd since in this case
Remark: If we get , setting and using Fermat's Little Theorem + The Binomial Theorem we get a proof of Wilson's Theorem.
Second Proof: The number of labeled trees with vertices such that a given set -you choose it- of labels correspond to leaves is , so in fact what is counting is the number of trees with no leaves, of course, if there are vertices this number will be 0.
2. We write : where , estimating this the rest follows.