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Source. Lemma 3 and equations (8)–(14), printed pp. 87–88 (PDF pp. 3–4). Let $\theta=11/10$, $I=\theta\log\theta-\theta+1$, and $c=I/4$ as in equation (9).

Statement. All but $o(x/(\log x)^c)$ positive integers $n\le x$, written

$$n=\prod_{i=1}^k p_i^{a_i},\qquad p_1<\cdots<p_k,$$

have an index $j$ such that

$$p_j>(\log x)^{10}\prod_{i<j}p_i^{a_i}. \tag{1}$$

The empty product at $j=1$ is one. In particular, for each such $n$ there is a proper divisor $d<n$ for which $n/d>1$ and every prime factor of $n/d$ exceeds $d(\log x)^{10}$.

Full proof

Put $X=\log x$ and $\ell=\log X$. Remove the integers $n\le x/X$; those divisible by $p^2$ for some prime $p>X$; and those with $\Omega(n)\ge\theta\ell$. Their total number is $o(x/X^c)$ by Lemma 2, equation (9), and $0<c<1$.

For a remaining $n$, let $r$ be the number of its prime factors at most $X$, allowing $r=0$, and set

$$d_0=\prod_{i\le r}p_i^{a_i},\qquad T_1=X^{10},\qquad T_2=X^{2\ell}.$$

Then $d_0\le X^{\Omega(n)}<T_2$, and all exponents with index greater than $r$ equal one. If $r=k$, then $n=d_0<T_2<x/X$ eventually, a contradiction. Thus at least one larger prime exists.

Suppose (1) fails for every index greater than $r$. With $A=T_1T_2$, the first such prime satisfies $p_{r+1}\le A$, and inductively

$$p_{r+i} \le T_1d_0\prod_{h<i}p_{r+h} \le A^{\,1+\sum_{h<i}2^{h-1}} =A^{2^{i-1}}\qquad(1\le i\le k-r). \tag{2}$$

The exponent in (2) is a power of two, not $2i-1$. Since $k\le\Omega(n)<\theta\ell$ and $\beta=\theta\log2<1$,

$$\log p_k \le 2^k\log A \le X^\beta(10\ell+2\ell^2).$$

Consequently

$$\log n \le \log T_2+(k-r)\log p_k \le2\ell^2+\theta\ell X^\beta(10\ell+2\ell^2) =o(X).$$

For large $x$ this implies $n<x^{1/2}$, contradicting $n>x/X$. Some index $j>r$ must satisfy (1). Take $d=\prod_{i<j}p_i^{a_i}$. Its cofactor contains $p_j$ and only larger primes, proving the final assertion.

Precision. The prime cutoff is taken as $p_i\le\log x$ in the small part, so possible equality causes no missing square case. The proof includes an empty small-prime part and explicitly rules out the all-small-prime case. The source's double-exponential recurrence is made explicit in (2); the displayed calculation also justifies the final little-oh bound without relying on extraction of its nested superscripts.

The proper-divisor requirement in the final assertion is essential for the maximal pairwise-coprime argument: allowing $d=n$ would give the cofactor one, whose empty prime support cannot be covered by that argument.