Assume is a polynomial with . Define

Compute the values of .

Assume is a non-negative integer (otherwise is undefined at ). Then, we make the following claim:

** Claim: ** The polynomial has derivatives at 0 given by the following

* Proof. * Since is a polynomial we may write,

Furthermore, since is given we know . Now, multiplying by we have

Next, we will use induction to prove that the th derivative of for is given by

for constants . Since the derivative is given by

for constants , we see that the formula holds for . Assume then that it holds for some ,

Then, taking the derivative of this we have,

Hence, the formula holds for all . But then, if we have

If then for all ; hence,