Communications engineering, call arrival rate, poisson
The key is actually that the results of the individual trials are independent of each other. If you get a successful "arrival" on the first trial, it doesn't affect what you get on the 2nd or later trials. This gives the binomial distribution, which is described by the probability mass function given in your question. It's easily derived:

You're making n trials. The result of each trial will be either an "arrival" or a "failure". The probability of failure in each trial is given by p.

The total number of (ordered) outcomes is \$2^n\$. For example, for 3 trials you could have aaa, aaf, afa, aff, faa, faf, ffa, or fff. The probability of each of these ordered results happening is \$p^k(1-p)^{(n-k)}\$, because if p is the probability of an arrival, then (1-p) must be the probability of failure.

So then you apply combinatorics. You've decided (or been told) you don't care about the order of the results, just how many arrivals. So out of the total \$2^n\$ ordered results, the number that meet your requirement is given by \$n \choose k\$, which is \$\frac{n!}{k!(n-k)!}\$.

Taken together, you have \$n\choose{}k\$ results that meet your requirements, each with probability \$p^k(1-p)^{(n-k)}\$, so the total probability is 

\${n\choose{}k} p^k(1-p)^{n-k}\$

which is the result you were looking for.