Simply put -since the sums are finite:

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

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- February 6th 2011, 08:00 AMPaulRS
- February 6th 2011, 09:23 AMUnbeatable0
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. - February 6th 2011, 11:56 AMchiph588@
- February 7th 2011, 04:50 AMPaulRS
My solutions:

**1.**

: Consider the set of functions from a set .*First proof*

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.

: 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.*Second Proof*

**2.**We write : where , estimating this the rest follows.