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Let $X_t$ count $t$-cliques, and put ${\mu }_t=\left(\genfrac{}{}{0ex}{}{n}{t}\right)p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Since ${\mu }_{r+1}\to 0$, the Markov inequality gives $\mathrm{Pr}(X_{r+1}>0)\to 0$. Any larger clique contains an $(r+1)$-clique, so it remains to prove $X_r>0$ with high probability.

We give the required growing-clique overlap bound explicitly. Set $\mu ={\mu }_r$ and, for $2\le s\le r$,The quantity $a_s$ is the probability that a uniformly chosen $r$-set overlaps a fixed $r$-set in exactly $s$ vertices. Two clique indicators share exactly $\left(\genfrac{}{}{0ex}{}{s}{2}\right)$ edge requirements, so their joint-to-product ratio is $p^{-\left(\genfrac{}{}{0ex}{}{s}{2}\right)}$. Overlap zero or one gives independence. Hence the overlap formula for the variance of a clique count impliesWe must bound this sum with $r$ allowed to grow.

Eventually ${\mu }_r\ge 1$ and ${\mu }_{r+1}\le 1$. Using $(n/t{)}^t\le (\genfrac{}{}{0ex}{}{n}{t})\le (en/t{)}^t$, these inequalities giveFor $2\le s\le r/2$, a union bound over the $s$ selected shared vertices gives $a_s\le \left(\genfrac{}{}{0ex}{}{r}{s}\right)(r/(n-r){)}^s$. Put $a=(s-1)/(r-1)<1/2$. Since $n\ge r^3$ and $n-r\ge n/2$,For each fixed $s$, the first bound tends to zero if $r$ stays bounded, because its power of $n$ is negative. If $r\to \infty$, the middle bound tends to zero because $a\to 0$. Thus each fixed-$s$ term tends to zero along arbitrary admissible sequences. The bounding series ${\sum }_{s\ge 2}(2e^2/s{)}^s$ converges. Splitting at a fixed large index, then taking $n\to \infty$ and the index to infinity, proves that the entire lower-half sum is $o(1)$.

For the upper half write $s=r-j$. The exact diagonal comparison isFor $1\le j\le r/2$, the second bound on $p$ givesThe last inequality uses $n\ge r^3$: the exponent of $n$ in the bracket is negative, so substituting $n=r^3$ bounds it above. Uniformly for this range of $j$, $r^{2(j+1)/r}\le 9j$. Indeed $r^{2/r}\le 3$, and concavity of $\log (3j)-(2j/r)\log r$, checked at $j=1$ and $j=r/2$, gives $r^{2j/r}\le 3j$. It follows thatThus $\mathrm{Var}{X}_r=o({\mu }^2)$. The second moment method gives $\mathrm{Pr}(X_r=0)\to 0$. Combining this with the absence of larger cliques provesThe restriction on $r$ is used in both overlap ranges; divergence of a clique count in a binomial random graph alone would not suffice for arbitrary growing $r$.