claim
$\forall \alpha, \beta, \gamma$ which are ordinals, $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$
proof
Let $\alpha = ord(X)$ $\beta = ord(Y)$ $\gamma = ord(Z)$ $\forall X,Y,Z$ which are finite-ordered set or $\Bbb N$.
$\alpha(\beta + \gamma) = ord(X \times(Y\cup Z))$ and
$ord(X \times(Y\cup Z)) = ord((X\times Y)\cup(Y\times Z))$ since
$X \times(Y\cup Z) = (X\times Y)\cup(Y\times Z)$.
Then $ord((X\times Y)\cup(Y\times Z))= ord(X\times Y) +ord(Y\times Z)$ since
$ord(A\cup B) = ord(A) + ord(B) \;\forall A,B$ which are finite well ordered set
Now, $ord(X\times Y) +ord(Y\times Z) = \alpha\beta + \alpha\gamma\;$ since
$ord(A\times B) = ord(A)ord(B) \;\forall A,B$ which are finite well ordered set.
Thus, $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$.