Turán's theorem proof I have a trouble with the following part of Turán's theorem. Some versions of the theorem are proved by exploiting lemma:
Turán graph $T_{k,n}$ is obtained by taking a set of $n$ vertices, dividing the vertices into $k$ classes in such a way that any two classes differ in size by at most one, and joining two vertices by an edge if and only if they are in different classes.

The number of edges in $T_{k,n}$ is larger than the number of edges $T_{k-1,n}$.

Why does this hold?
 A: Let $e(G)$ denote the number of edges in a graph $G.$
Recall that $K_{n_1,\dots,n_k}$ is the complete $k$-partite graph with vertex set $V=V_1\cup\cdots\cup V_k$ where the partite sets $V_i$ are pairwise disjoint, and two vertices are joined by an edge just in case they are in different partite sets.
$K_{n_1,\dots,n_k}$ is the Turán graph $T_{k,n}$ when $|n_i-n_j|\le1$ for all $i,j.$
Lemma. Let $n,k\in\mathbb N,\ k\le n.$ Among all graphs $K_{n_1,\dots,n_k}$ with $n_1+\cdots+n_k=n,$ the number of edges $e(K_{n_1,\dots,n_k})$ is maximized when $K_{n_1,\dots,n_k}$ is the Turán graph $T_{k,n}$.
Proof. Let $G=K_{n_1,\dots,n_k}$ be the graph with the maximum number of edges, and assume for a contradiction that $G$ is not the Turán graph. Then we have $|n_i-n_j|\gt1$ for some $i,j;$ without loss of generality, we may assume that $n_1-n_2\gt1.$ Then, since $(n_1-1)(n_2+1)\gt n_1n_2,$ we have $e(K_{n_1-1,n_2+1,n_3,\dots,n_k})\gt e(G),$ contradicting the masimality of $G.$
Theorem. If $2\le k\le n,$ then $e(T_{k-1,n})\lt e(T_{k,n}).$
Proof. Let $T_{k-1,n}=K_{n_1,\dots,n_{k-1}}$ where $n_1\ge n_2\ge\cdots\ge n_{k-1}.$ Then
$$e(T_{k-1,n})=e(K_{n_1,n_2,\dots,n_{k-1}})\lt e(K_{1,n_1-1,n_2,\dots,n_{k-1}})\le e(T_{k,n}),$$
the last inequality from the Lemma.
For the penultimate inequality, observe that, if $G_1$ is a complete $(k-1)$-partite graph with partite sets $V_1,V_2,\dots,V_{k-1},$ and if we partition $V_1$ into two disjoint nonempty sets $V_1',V_1''$ and let $G_2$ be the complete $k$-partite graph with partite sets $V_1',V_1'',V_2,\dots,V_{k-1},$ then $e(G_1)\lt e(G_2),$ because $G_2$ has all the edges of $G_1$ plus additional edges between $V_1'$ and $V_1''.$
