We will prove that
. Then
for all
, and thus if
,
.
We first prove that if is a prime, then
.
Consider the set
,
where
is the
smallest value such that
. (Such a value must
exist since the
number of possible values for
is limited by the numbers less
than
. Thus at some point two of these powers, say
and
we must have
(I assume
). Thus
must be a multiple of p. Since by assumption
does not have
as a factor, neither does
, and thus
must
have
as a factor. This implies that
.)
The set of all powers of is a subgroup of the group of
all numbers from 1 to
under
multiplication. We now rely on the theorem of group theory that the
size of a subgroup always evenly divides the size of the group itself. That
means that r must evenly divide
, and thus
Note that this
theorem is true, even if
is greater than
. Thus for any
,
for some integer
.
We now show that
.
Using
and using the
binomial theorem, we have
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