Finite Connected Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step/Proof 1

Proof
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.

Take any node $v$ of $T_{k + 1}$ of degree $1$.

Such a node exists from Finite Tree has Leaf Nodes.

Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree.

By Connected Subgraph of Tree is Tree, $T_k$ is itself a tree.

The order of $T_k$ is $k$, and it has one less edge than $T_{k + 1}$ by definition.

We have that $T_k$ has $k - 1$ edges.

So $T_{k + 1}$ must have $k$ edges.