Let $T$ be a finite tree with $V(T)$ vertices and $E(T)$ edges. Then $V(T)-E(T)=1$.
I am really confused with this. I am not sure how to go about proving it. I can (and should?) use Euler's theorem that states that $V(T)-E(T)+F(T)=2$. Would appreciate some help on this one.
No need to use Euler's Theorem.
You could prove this by weak induction over the number of vertices:
Base: V =1. Then we have 1 vertex by itself which is a tree by itself, having 0 edges. So $V-E=1-0=1$ Check!
Step: Suppose for any tree $T$ with $V = n$ vertices and $E$ edges we have $V=E-1$. Now consider any tree $T'$ with $V'=n+1$ vertices and $E'$ edges. Pick one of the vertices, and start a path following edges without repeating any edge until you can no longer go any further. Since the tree contains no cycles, this means that the vertex you end up with must have been a vertex with an odd degree (since otherwise it would not be the end of the path), and in fact it has to have degree 1, because without any cycles you can't visit any vertex twice along this path. So, if you remove that last vertex and edge, you end up with a tree with $n$ vertices which by inductive hypothesis has $n-1$ edges. But that means that the original tree $T'$ had $n$ edges, meaning that $V' - E' = (n+1) - n = 1$. Check!
You can also prove it using strong induction:
Pick any tree $T$ with $V$ vertices and $E$ edges. Pick any vertex $p$ in this tree. If $V=1$, then there are no vertices other than $p$, so there are no edges either, so $V - E = 1-0=1$. If $V>1$ then, it being a tree, $p$ must share an edge with $n \ge 1$ other vertices $p_1,...p_n$. If you remove $p$ and its $n$ edges from the tree, then you end up with a graph consisting of $n$ disconnected trees, because there cannot be another path from any $p_i$ to $p_j$, for otherwise there would have been a cycle in the original graph, and hence it would not have been a tree. Say that these $n$ trees have $V_i$ vertices and $E_i$ edges respectively. Since $V_i < V$, we can apply the inductive hypothesis and say that $V_i - E_i = 1$. We also have:
$V = V_1+V_2+...+V_n+1$ (we removed $p$)
$E = E_1+E_2+...+E_n+n$ (we removed $n$ edges)
So:
$$V - E = V_1+V_2+...+V_n+1 - (E_1+E_2+...+E_n+n) =$$
$$ (V_1 - E_1) + (V_2 - E_2) + ... + (V_n - E_n) + 1 - n =$$
$$ n\cdot 1 + 1 - n = n + 1 - n = 1$$
