Well ordering of $N^N$

Question

Is $N^N$ a well order under dictionary ordering?
I realize that the set $N/(0,0,0,0,0,....)$ does not contains a least element so it is not well ordered but why are we considering $0$ when it is not in $N$.
Why can't $(1,1,1,1,.....)$ be least in $N^N$ when $0$ is not even in $N$.

Answer 1

Whether you consider 0 to be in $\mathbb{N}$ or not is irrelevant. You always get the same problem.

For example the sequence $(2,1, \cdots ),(1,2,1, \cdots ), (1,1,2,1, \cdots ), \cdots $ is strictly decreasing, which contradicts the well ordering and we did not even need to use 0.