Here is one way to show that `(nm - 1)` breaks are sufficient. (It does not, however, demonstrate that `(nm - 1)` breaks are *necessary*.)

The block of chocolate has `(n-1)` and `(m-1)` 'score lines' on each side, along which it can be broken. By making `(n-1)` breaks we can produce `n` long and thin pieces, of size `1 x m`. We then need to make `(m-1)` breaks on each of these long and thin pieces to get them down to `1 x 1` size. ie, `n(m-1)` more breaks.

This gives us a total of `(n-1) + n(m-1)` breaks.

`(n-1) + n(m-1)`

`= n - 1 + nm - n`

`= -1 + nm`

`= nm - 1`

QED