I am working on my number theory assignment and we have this theory that states that
sum of Euler's totient function of all positive divisor equals the number itself
this is just a little back ground, my question is that what is the total number of positive divisor of a number n $$ n = p_1^{\alpha 1} p_2^{\alpha2}...p_k^{\alpha k} (prime factorization)$$ my argument is that number of positive divisor is $$(\alpha1+1)(\alpha2+1)...(\alpha k+1)$$ I am using combinatorics, since this is the number of ways to choose subsets out of the prime factorization with no duplicates (and because prime factorization is unique, if I multiply primes in these subsets, they should give me unique divisors?), however I haven't done combinatorics for a while, I need to verify the correctness of this equation...so far it works for primes: $$n = p$$ $$1+1 = 2$$