Rolling dice probability

Here's my thought. Check if it is correct. I also want to know if there's a simpler method.

Given any $i$ numbers from $1$ to $k$ , let $w(i)$ be the number of scenarios under which all and only those $i$ numbers appear in our $n$ trials. Then we have $w(1)=$ 1, and $w(i)={i}^{n}-\sum_{j=1}^{i-1}{i \choose j}w(j)$ .

By this definition, there are exactly ${k \choose i}w(i)$ scenarios under which exactly $i$ different numbers from $1$ to $k$ appear.

Hence the required probability $p=({k}^{n}-\sum_{i=1}^{k-1}{k \choose i}w(i))/{m}^{n}$ .