A proof that all even perfect numbers are a power of two times a Mersenne prime  (From the Prime Pages' list of proofs)
 Over 2300 years ago Euclid proved that If 2k-1 is a prime number (it would be a Mersenne prime), then 2k-1(2k-1) is a perfect number. A few hundred years ago Euler proved the converse (that every even perfect number has this form).  It is still unknown if there are any odd perfect numbers (but if there are, they are large and have many prime factors). Theorem: If 2k-1 is a prime number, then 2k-1(2k-1) is a perfect number and every even perfect number has this form. Proof:  Suppose first that  p = 2k-1 is a prime number, and set n = 2k-1(2k-1).  To show n is perfect we need only show sigma(n) = 2n.  Since sigma is multiplicative and sigma(p) = p+1 = 2k, we know sigma(n) = sigma(2k-1).sigma(p) =  (2k-1)2k = 2n. This shows that n is a perfect number.     On the other hand, suppose n is any even perfect number and write n as 2k-1m where m is an odd integer and k>2.  Again sigma is multiplicative so sigma(2k-1m) = sigma(2k-1).sigma(m) = (2k-1).sigma(m). Since n is perfect we also know that sigma(n) = 2n = 2km. Together these two criteria give 2km = (2k-1).sigma(m), so 2k-1 divides 2km hence 2k-1 divides m, say m = (2k-1)M.  Now substitute this back into the equation above and divide by 2k-1 to get 2kM = sigma(m).  Since m and M are both divisors of m we know that 2kM = sigma(m) > m + M = 2kM, so sigma(m) = m + M.  This means that m is prime and its only two divisors are itself (m) and one (M).  Thus m = 2k-1 is a prime and we have prove that the number n has the prescribed form.
 Another prime page by Chris K. Caldwell