If $n$ is a product of primes, what is the number of divisors? Let $n=p_1p_2...p_k$ Then the number of divisors is what?
I assumed it was $1+ \binom k1+ \binom k2 + \binom k3 + ... + \binom kk=2^k$
Is this correct?
Prove that the number of divisors is odd $\iff$ n is a perfect square
 A: It depends. Your answer 
$$ \sum_{m = 0}^k \binom km = 2^k $$ 
is correct, if all $p_i$ are distinct. Note that this makes a difference, $n = 2 \cdot 2$ has $(1,2,4)$ as divisors (namely 3), where $6 = 2 \cdot 3$ has $(1,2,3,6)$, so $4 = 2^2$. 
In general, if we have 
$$ n = \prod_i p_i^{\alpha_i} $$
with distinct primes $p_i$, for each $i$, we can choose $0, \ldots, \alpha_i$ factors $p_i$ in a divisor of $n$, so alltogether we have 
$$ \prod_i (\alpha_i + 1) $$
divisors. Note that is number is odd iff all $\alpha_i$ are even. But then 
$$ n = \prod_i p_i^{\alpha_i} = \prod_i p_i^{2\beta_i} = \left(\prod_i p_i^{\beta_i}\right)^2. $$
A: You can write it down easier: Suppose, that $n=p_1^{l_1} p_2^{l_2} ... p_n^{l_n}$, where $l_i \in \mathbb{N_+}$. In that case, how you can get a divisor? If you take $p_1$, that is a divisor, if you take $p_1^2$, that is divisor too, if you take $p_1 p_2$, etc. You can see, that you have $(l_1+1) * (l_2+1) *... *(l_n+1)$, since with every prime, you have $l_i$ possibilities, how to choose it, and the $+1$ is because you can choose not to choose it.
An example to make it clear: $12=2^2*3$, therefore: $l_1=2, l_2=1$. Number of divisors: $(2+1)*(1+1)=6$, since if $l_1=0, l_2=0$, it is $1$, if $l_1=1, l_2=0$, it is $2$, if $l_1=2, l_2=0$, it is $4$, if $l_1=0, l_2=1$, it is $3$, if $l_1=1, l_2=1$, it is $6$, if $l_1=2, l_2=1$, it is $12$. You got all of them.
For the second part: $m$ is a perfect square iff $m$ has an odd number of divisors?
A number is a perfect square if and only if it has odd number of positive divisors
