Prime divisor greater than Sylow index is Sylow-unique

Statement
Suppose $$p$$ is a prime number and $$m$$ is a natural number less than $$p$$. Suppose:

$$n = p^km$$

for some nonnegative integer $$k$$. Then, $$p$$ is a fact about::Sylow-unique prime divisor of $$n$$.

Facts used

 * 1) uses::congruence condition on Sylow numbers
 * 2) uses::divisibility condition on Sylow numbers

Related facts

 * Order is product of Mersenne prime and one more implies normal Sylow subgroup

Proof
Given: A prime $$p$$, a natural number $$m < p$$, a group $$G$$ of order $$n = p^km$$ for $$k \ge 0$$.

To prove: $$n_p = 1$$ in $$G$$.

Proof: By fact (1), $$n_p \equiv 1 \mod p$$. By fact (2), we also have $$n_p | m$$, Thus, $$n_p \le m$$.

Combining the conditions $$n_p \equiv 1 \mod p$$ and $$n_p \le m$$ yields that $$n_p = 1$$.