I could use some help with proving the following irreducibility criterion. (It came up in class and got me interested.)

Let p be a prime. For an integer $n = p^k n_0$, where p doesn't divide $n_0$, set: $e_p(n) = k$. Let $f(x) = a_n x^n + \cdots + a_1 x + a_0$ be a polynomial with integer coefficients. If:


$e_p(a_n) = 0$, 
$e_p(a_i) \geq n - i$, where $i = 1, 2, \ldots, n-1$,
$e_p(a_0) = n - 1$,


then f is irreducible over the rationals. 

Reducing mod p and mimicking the proof of Eisenstein's criterion doesn't cut it (I think). I also tried playing with reduction mod $p^k$, but got stuck since $Z_{p^k}[X]$ is not a UFD.

Also, does this criterion has a name?

I think this question confuses a lot of people because there's a lack of intuitive context -- I'll try to supply that.

Suppose there is a birthday party to which all of the girls (and none of the boys) in a small town are invited. If you run into a mother who has dropped off a kid at this birthday party and who has two children, the chance that she has two girls is $1/3$. Why? $3/4$ of the mothers with two children will have a daughter at the birthday party, the ones with two girls ($1/4$ of the total mothers with two children) and the ones with one girl and one boy ($1/2$ of the total mothers with two children). Out of these $3/4$ of the mothers, $1/3$ have two girls.

On the other hand, if the birthday party is only for fifth-grade girls, you get a different answer. Assuming there are no siblings who are both in the fifth grade, the answer in this case is $1/2$. The child in fifth grade is a girl, but the other child has probability $1/2$ of being a girl. Situations of this kind arise in real life much more commonly than situations of the other kind, so the answer of $1/3$ is quite nonintuitive.
