We will start with the following lemma.
Lemma 1. There exists a diagonal that has two equal numbers for any $2 \times 2$ square.
Proof. Let $X$ be the greatest number in $2 \times 2$ square. Let his neighbors in this $2 \times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if $A = B$, the lemma is proven. Suppose $A > B$. Then $A + X = k! > B + X = m! > 1$, thus $k > m$. It is also clear that $m > 1, k > 2$. But then
$$2(B + X) = 2m! > 2X \ge A + X = k! \ge k \cdot m! > 2m!,$$
that leads to a contradiction. This finishes the proof of Lemma 1.

By contradiction, suppose $a, b, c, d, e, f, g, h, i$ are in the table (Fig. 5). By Lemma 1 for the square that consists of $a, b, d, e$ either $b = d$ or $a = e$. In the first case, we will use Lemma 1 for the square with $d, e, g, h$, and then for $b, c, e, f$. Thus, either $b = d = h$, or $b = d = f$, or $g = e = c$.

In the second case, we will use Lemma 1 for $e, f, h, i$, and then for $b, c, e, f$. Thus, either $a = e = i$ or $a = e = c$, or $h = f = b$. In all cases we obtain three equal numbers.