# A $k$-regular tree is unique

I am trying to prove that a $k$-regular tree is unique upto isomorphism. Clearly except for the trivial cases $K_1$ or $K_2$ any such tree must be infinite, for if it had $n$ vertices then $k=\frac{2(n-1)}{n}$ (by the handshaking lemma and the fact that in a tree the number of edges is one less then the number of vertices). Since $(n-1,n)=1$ so $n\mid 2$ which forces the trivial cases.

Furthermore it is evident that such a tree $T$ may be constructed by taking a vertex $v$, giving it $k$-neighbors, then giving each of the neighbors a further $k-1$ neighbors, and so on. What is not evident to me is as to why every $k$-regular tree will be isomorphic to $T$. Can I simply construct an isomorphism like this: Pick any vertex $x$. Correspond it to $v$. Arbitrarily correspond each of the $k$ neighbors of $x$ to the neighbors of $v$ in a one to one fashion. Continue doing so. Is this is a valid way to describe an isomorphism?

Let G be another k-regular tree. choose a vertex in $G$ and map it to the 'base' of $T$. Map its neighbours to the neighbours of the base. Map the neighbours of those neighbours to the neighbours of the neighbours. Repeat until you have an isomorphism!