fundamental theorem of arithmetics:
Every natural #n > 1# can be expressed as a product of primes.
Two factorizations are identical, "apart from" the order.
Unicity:
#n = p_1 ^ (a_1) times ... times p_k ^(a_k) = q_1 ^ (b_1) times ... times q_j ^(b_j)#
Every #p_i# divides #q_1 ^ (b_1) times ... times q_j ^(b_j)#
"so that every #p# is a #q#"
Again: Every #q_i# divides #p_1 ^ (a_1) times ... times p_k ^(a_k)#
"so that every #q# is a #p#"
"Hence" #k = j# and #p_i = q_i#, forall #i#.
Suppose by contradiction that #a_i > b_i#
#n / p_i ^ (b_i) = p_1 ^ (a_1) times ... times p_i ^(a_i - b_i) times ... times p_k ^(a_k) = q_1 ^ (b_1) times ... times q_(i - 1)^(b_(i - 1)) times 1 times q_(i + 1)^(b_(i + 1)) times ... times q_j ^(b_j)#
Left side is divisible by #p_i#, right side is not.
Again:
Suppose by contradiction that #b_i > a_i#
#n / p_i ^ (a_i) = q_1 ^ (b_1) times ... times q_i ^(b_i - a_i) times ... times q_k ^(b_j) = p_1 ^ (a_1) times ... times p_(i - 1)^(a_(i - 1)) times 1 times p_(i + 1)^(a_(i + 1)) times ... times p_j ^(a_k)#
Left side is divisible by #q_i#, right side is not.
Therefore #a_i = b_i#.
