How many ways are there to distribute 8 teachers to 4 schools where each school must get at least 1 teacher? Additional details: the teachers are considered distinct from one another.
So here is what I thought:
1) Choose four teachers to go to each one of the schools: $\binom{8}{4}\cdot4!$
2) For each of those situations, distribute the other 4 teachers to the 4 schools: $4^4$
So total: $\binom{8}{4}\cdot4!\cdot4^4$
However, I am almost 100% sure that I am over-counting but can't quite put my finger on it. Any help would be appreciated.

Another way is using exponential generating functions.
The EGF of this problem is:
$\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+
\frac{x^7}{7!}+\frac{x^8}{8!}\right)^4$
to solve the problem we can use 
$(e^x-1)^4$
for that expression. This expands out to
$ 1-4 e^x+6 e^{2 x}-4 e^{3 x}+e^{4 x}$
this can expanded by hand or taken over to Wolfram Alpha. We get:
$x^4+2 x^5+\frac{13 x^6}{6}+\frac{5 x^7}{3}+
\frac{81 x^8}{80}+...+$
The coefficient of x^8 is the one we want and we multiply that by 8!
$\dfrac{81}{80}\cdot 8! = 40824$
That is the number of ways.
