Difference Operator and the falling factorial power of k

I have been able to show by algebraic manipulations that the when one applies the difference operator to "x to the n falling" one finds equality with n times x to the (n-1) falling.

I know that I have to use this now, but no matter how I try to manipulate the following expression, I can't get the result.

Prove that the sum from k=0 to (m-1) of k to the n falling is equal to m to the (n+1) falling all divided by (n+1)

I've realized that values where n is greater than or equal to m results in a sum of zero, and when n < m, the first n terms of the sum are zero. However, I can't seem to find where to best use the first statement or where I should be focusing my attention.