It easy to see that each of the prime numbers ($p_{i}$) has to be a divisor of exactly one of $\alpha_{i}$, so therefore the $\alpha_{i}$ which it divides can be one of $k$ integers. Since there are $k$ prime numbers ($p_{1}, ..., p_{k}$), we get that the solution to the problem is $k^k$.

No, the $\alpha_i$ does not have to be a permutation of the $p_i$. For example, take $k=2,p_1=2,p_2=3$; then $(\alpha_1,\alpha_2)$ can be $(1,6)$. Here's my solution (hopefully it is correct):