Does it work for some other larger number too?

Since there is only one factor of 7, we immediately know that must be the answer (if one exists), because 7 could not belong to either group. Verification is trivial.

The condition is, the product of the numbers before the number $\text{\_\_\_}$ is the same as the product of the numbers after that number.

Only $\color{#3D99F6} \boxed{7}$ satisfies the condition.

$\large \color{#D61F06} \Rightarrow \color{#3D99F6}1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = \color{#20A900}5040$

$\large \color{#D61F06} \Rightarrow \color{#3D99F6} 7 \times 8 \times 9 \times 10 = \color{#20A900}5040$

Hence the product of the numbers before the number $\color{#3D99F6} \boxed{7}$ is the same as the product of the numbers after that number.

The product of the numbers from 1 to 10 is 10!. We let the number fit to be in the blank be n. As both the products of the number before n and after n are equal (and an integer), 10!/n is a square number.
The only number satisfying this is 7
We then check is 7 is correct. 6!=720= $8*9*10$

Let the product be $\text{p} = \displaystyle\prod_{k=1}^{10}k = 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$ here $4,6,8,9,10$ are composite numbers that has the prime factor of $2,3,5$ and there doesn't exist any number that can be expressed as the prime factor of $7$ . So, $7$ is the intermediate number in which the product of number before it and after it is equal. $\text{p} =\displaystyle\prod_{k=1}^{10} =1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10 =2^8\times 3^4\times 5^2\times 7^1$ . The power to the prime are in geometric progression for $\displaystyle\prod_{k=1}^{10}k$ however,for $k>10$ doesn't form G.P which fails for other such numbers.