Let $K_{n}$ is the complete graph on $n$ vertices and $T_{m}$ is a tree on $m$ vertices
How do I show that $R(K_{n} , T_{m}) = (n-1)(m-1) + 1$?
Here $R(G,H)$ is the minimum $t \in \mathbb{N}$ such that every $2$ coloring of edges of $K_{t}$ yields a red $G$ or a blue $H$, for graphs $G,H$. I was thinking of using induction but I am stuck.
The original source of this result seems to be a one-page note of Václav Chvátal, Tree-complete graph Ramsey numbers, J. Graph Theory 1 (1977), 93. I haven't seen Chvátal's note, but I suppose his argument is something like the following.
The inequality $R(K_n,T_m)\ge(n-1)(m-1)+1$ is trivial; there is an obvious red/blue coloring of the edges of a complete graph of order $(n-1)(m-1)$ with no red $K_n$ and no blue tree of order $m$.
I have to show that $R(K_n,T_m)\le(n-1)(m-1)+1.$ Let me restate it this way:
Theorem. If $G$ is a graph of order $r=(n-1)(m-1)+1$ with independence number $\alpha(G)\lt n$, then $G$ contains an isomorphic copy of every tree of order $m$.
We can assume that $n\gt1$. Since $$\chi(G)\ge\frac r{\alpha(G)}\ge\frac r{n-1}\gt m-1,$$ we have $\chi(G)\ge m$, and so $G$ contains a minimal $m$-chromatic subgraph $H$, which has minimum degree $\delta(H)\ge m-1$. Thus it will suffice to prove the following:
Lemma. If $H$ is a graph with minimum degree $\delta(H)\ge m-1$, then $H$ contains an isomorphic copy of every tree of order $m$.
The lemma is easily proved by induction on $m$. Maybe it's a theorem or exercise in your graph theory textbook; e.g., it's Proposition 2.1.8 on p. 70 of Douglas B. West's Introduction to Graph Theory, Second Edition.