From (
), one can see that 
and hence 
is
dependent on the choice of an arbitrary term 
. However, the product 
is invariant under choice of
any of the arbitrary terms 
,
which we will now demonstrate by induction for the
non-degenerate case. The extension to the degenerate case is straightforward.
We need to prove that the Taylor series coefficients of the
product are invariant, i.e. that 
is invariant
for every n. This is clearly true for n=0. Assume that
this is true for all n less than some whole number N. Then we find
by use of (). The first term on the right
hand side is invariant by hypothesis. Now consider the second
term, and suppose that 
is invariant for all n less than N,
as it is clearly true for n=0. Then
The first term of the right hand side is invariant by hypothesis, so we only need consider the second term
Clearly, this term is invariant by hypothesis, so
is invariant.