Turán's theorem with equality: $\mathrm{ex}(n,K_{r+1})=e(T_{n,r})$, and $T_{n,r}$ is the unique extremal graph

The graph $T_{n,r}$ is $K_{r+1}$-free. Zykov symmetrisation sends an extremal graph to a complete $k$-partite graph with $k\le r$ and the same edge count; adding empty parts makes it complete $r$-partite, so balancing bounds its edges by $e(T_{n,r})$. Hence the displayed extremal number is exact.

For uniqueness, induct on $r$. At $r=1$, a $K_2$-free graph is edgeless and equals $T_{n,1}$. The case $n=0$ is also immediate. Assume $r\ge 2$, $n\ge 1$, and rigidity for $r-1$, and let $G$ attain $e(T_{n,r})$. Choose a vertex $v$ of maximum degree $d$, put $A=N(v)$ and $B=V(G)\setminus A$. Then $G[A]$ is $K_r$-free and $e(G)\le e(G[A])+{\sum }_{b\in B}d(b)\le e(T_{d,r-1})+(n-d)d$. The last expression is the edge count of a complete $r$-partite graph whose one part has size $n-d$ and whose remaining parts are balanced on $d$ vertices, so balancing makes it at most $e(T_{n,r})$.

Equality for $G$ forces equality throughout step 1.2. The first inequality forces $G[B]$ to have no edge, the degree inequality forces every $b\in B$ to have degree $d$, and $\mathrm{\mid }A\mathrm{\mid }=d$ then forces every vertex of $B$ to be adjacent to every vertex of $A$. Inductive rigidity gives $G[A]\cong T_{d,r-1}$, and balancing equality makes the resulting $r$ part sizes differ by at most $1$. Thus $G\cong T_{n,r}$.

Conversely $T_{n,r}$ is $K_{r+1}$-free and has the extremal edge count by step 1.1. The induction therefore proves both directions of the equality characterization.

Steps 1.1-3.1 prove the exact formula and uniqueness for every $n$, including $n<r$ and $n=0$.