If $M_1$ and $M_2$ are well ordered then $ M = M_1 \times M_2$ is well ordered too.
with the orders $$ (a_1,b_1) \leq (a_2,b_2) \ iff \ b_1 \leq b_2 \ in \ M_2 \ and \ a_1 \leq a_2 \ in \ M_1 \ ,\ b_1=b_2$$
I prove that this is a total order, I prove trichotomy, transitive, but how can I prove that there existe a first element in M.
I tried this by contradiction if $(a_0,b_0) \in M$ is first element but I don't know how continue this idea.
Someone can help me please. Thanks for your time and help.
The "and" in your definition should be "or", in which case it works. Let $a_0$ be the first element in $M_1$ and $b_0$ be the first element in $M_2$. You are guaranteed they exist because $M_1,M_2$ are well ordered. Then $(a_0,b_0)$ is the first element. You can use your definition to show it is less than ore equal to any element of $M$. For a well order every subset of $M$ must have a least element, but you take any collection, take the elements with the smallest second items, and take the one of those that has the smallest first item.