And we are done since
Now a couple of problems:
1. Show that:
2. Show that: for all
Here is the number of positive integers dividing .
For the case to work, we write
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,
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.