In how many ways can eight different balls be distributed among 4 kids, s.t each gets at least one.

my approach: 
so I read a similar question here and figured out I should do this:
$$4^8 - \binom{4}{1}3^8 + \binom{4}{2}2^8 - \binom{4}{3}1^8$$

but then I thought about this way : I will make sure first that each child gets one ball , so:  

$8$ options for the first one  

$7$ for the second  

$6$ for the third  

$5$ for the fourth  

and then I have $4^4$ ways to distribute the left $4$ balls among the $4$ children. total $$(8 \cdot 7 \cdot 6 \cdot 5) \cdot 4^4$$

unfortunately the solution is not the same, so one of those approaches is wrong, but I cannot figure out which one. 

Please help