Axiomatising Wellorder It may be of use to recall what a strict total order is; namely a binary relation satisfying irreflexivity, transitivity and totality, as formalised below: 
$$\forall x(\neg P_{1}(x,x))$$
$$\forall x \forall y \forall z((P_{1}(x,y) \land P_{1}(y,z)) \to P_{1}(x,z))$$
$$\forall x \forall y (P_{1}(x,y) \lor x \dot{=}y \lor P_{1}(y,z))$$
Furthermore, a well-order is a strict total order, $\mathcal{A}=\langle A; <;-\rangle $ for  which there is no infinite sequence $\{x_{j}\}_{j \in \mathbb{N}}$ of elements in $A$, such that $x_{j+1}<x_{j}$ for all $j \in\mathbb{N}$.  
Now,  what I want to show is:

Well-order is not axiomatizable. 

My idea so far has been to suppose that $\Gamma$ axiomatizes the class of well-orderings, add countably many constants $c_{j}$ and show that $\Gamma \cup \{c_{j+1}<c_{j}:j\in\mathbb{N}\}$ has a model. However, I find it difficult to follow through on the last step, namely show that $\Gamma \cup \{c_{j+1}<c_{j}:j\in\mathbb{N}\}$, which is why I ask for help from this forum.    

Another method, which may be slightly simpler, is to use ultraproducts and Los theorem, if you have studied about them.
Note that $\cal N=\langle\Bbb N,<\rangle$ is a strict well-order. Suppose that it was axiomatizable then every ultraproduct of well-orders would be a well-order. Consider any free ultrafilter on $\Bbb N$, $\cal U$ then the structure: $$\cal M=N^\Bbb N/U$$
Should be a well-order as well, since by Los theorem both $\cal N$ and $\cal M$ have the same first-order theory. However it is not hard to find a decreasing sequence in $\cal M$. Such sequence, for example, is letting $f_0$ to be the identity function, and $f_{n+1}(k)=f_n(k-1)$ (and $0$ for $k=0$). 
