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
$$2(B + X) = 2m! > 2X \geq A + X = k! \geq k \cdot m! > 2m!,$$
that leads to a contradiction. This finishes the proof of Lemma 1.

By contradiction, let the table consist of the numbers $a_1, a_2, ..., a_{16}$, that satisfy the conditions (Fig. 7).
Use Lemma 1 for a square with $a_1, a_2, a_5, a_6$. Without loss of generality, let $a_2 = a_5$. Use Lemma 1 for squares with $a_2, a_3, a_6, a_7$ and $a_5, a_6, a_9, a_{10}$.

Case I. $a_2 = a_7$ or $a_5 = a_{10}$. Since these cases are similar, let $a_2 = a_5 = a_{10} = x$.

<table><tr><td>$a_1$</td><td>$x$</td><td>$a_3$</td><td>$a_4$</td></tr><tr><td>$x$</td><td>$a_6$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$a_9$</td><td>$x$</td><td>$a_{11}$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$a_{14}$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>
Fig. 8

From Lemma 1 for the square $a_9$, $a_{10} = x, a_{13}, a_{14}$, if $a_{13} = x$, then we have four equal numbers that lead to a contradiction. Then $a_9 = a_{14}$. Similarly, from Lemma 1 for a square with $a_{10} = x, a_{11}, a_{14}, a_{15}$ we have that $a_9 = a_{14} = a_{11} = y$

<table><tr><td>$a_1$</td><td>$x$</td><td>$a_3$</td><td>$a_4$</td></tr><tr><td>$x$</td><td>$a_6$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$y$</td><td>$x$</td><td>$y$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$y$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>
Fig. 9

It suffices to use the Lemma 1 for a square with $a_6, a_7, a_{10} = x, a_{11} = y$ thus, the table has either four numbers $x$ or four numbers $y$. The contradiction completes the proof.

Case II. $a_3 = a_6 = a_9 = t$.

<table><tr><td>$a_1$</td><td>$a_2$</td><td>$t$</td><td>$a_4$</td></tr><tr><td>$a_5$</td><td>$t$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$t$</td><td>$a_{10}$</td><td>$a_{11}$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$a_{14}$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>
Fig. 10

By assumption, there is no other number $t$ among the rest of the values, so by Lemma 1 the following holds: $a_4 = a_7 = a_{10} = a_{13}$, that leads to a contradiction and completes the proof.