Edge exchange property of two Minimum Spanning Trees
Here is a proof. Let $V$ be the vertices of $G$. $V$ are also the vertices of $T$ and the vertices of $T'$.
If $e$ is deleted from $T$, we will get two trees. Let the vertices of these two trees be $(V_1, V_2)$, which is a cut of $V$.
Since $T'$ is a tree, if we add $e$ to it, we will obtain a cycle. Since that cycle crosses $(V_1, V_2)$ at $e$, it must cross $(V_1, V_2)$ at another edge, say, $e'\in T'$.

Since $T \setminus \{e\}\cup \{e'\}$ is a spanning tree of $G$ and $T$ is an MST,  $\text{weight}(e) \le \text{weight}(e')$.
Since $T' \setminus \{e'\}\cup \{e\}$ is a spanning tree of $G$ and $T'$ is an MST,  $\text{weight}(e') \le \text{weight}(e)$.

So, $\text{weight}(e) = \text{weight}(e')$.
$T \setminus \{e\}\cup \{e'\}$, which is a spanning tree of the same weight as $T$,  must be an MST of $G$. $\quad\checkmark$