As there have to be atleast $k$ balls in each bag, so firstly put $k$ balls in each bag i.e $\left(k*n\right)$ balls.
Then now we have total $\left(m-k*n\right)$ balls remaining.
We can use balls $\&$ sticks method now $!$
$n$ bags$= n$ variables, they need to equal to $\left(m-k*n\right)$, no restrictions on how many balls in each bag $!$
$x_{1}+x_{2}+\cdots+x_{n}= \left ( m-k*n \right ),x_1,x_2\cdots x_n>=0.$
So when we solve it
We get
$C(m - k*n + n - 1, n-1 ) = C(m - k*n + n - 1, m- k*n )$