The complete problem is the following:
You have a set of numbers of the form M = { 7, 77, 777, …, 77…7 (n times) }.
Can it be that you have two disjoint sets A and B such as $A \cup B = M$ have the property that the product of A's elements equals the product of B's elements?
Note: it's from a collection of problems for the 6th grade but I don’t mind of any solution.
I apologize for posting another answer, but apparently everyone, including me, missed a very simple proof why the product is never a square, which is indeed suitable for children: just consider everything modulo $4$. If the number ends in $77$ it is congruent to $1$ modulo $4$, and $7 \equiv 3\pmod 4$, and $3$ is not a square mod $4$.
I will show that even the product of the first $n-1$ terms of the set is not divisible by the $n$'th term; this is clearly enough since it has to go to one of the subsets. If $n = 1, 2, \ldots , 6$, then you can check it directly. So, assume that $n > 6$.
By Zsigmondy's Theorem $10^n - 1$ has a prime divisor $p$ that does not divide $10^k - 1$ for all $k < n$. I claim that $p > 7$ and that $p$ divides $77\ldots 7$ ($n$ sevens). Indeed, $p$ is clearly not $2$ and not $5$ since $10^n-1$ is coprime to $2$ and $5$. We also know that $p$ is not $3$ and not $7$ because $10^6-1$ is divisible by $3$ and $7$, and we assumed that $n > 6$. So, $p > 7$.
That $p$ divides $77\ldots 7$ ($n$ sevens) is also clear now because it is $7\frac{10^n-1}{9}$ and $p > 7$, so it still divides this number. Similarly, it does not divide any of the numbers $77\ldots 7$ (with $k$ sevens) because, once again, $p > 7$. So, our number does not divide the product of all other numbers because it is divisible by $p$, and all of the other ones aren't.
A short solution of the original problem. Denote $\underbrace{1\dots 1}_{k\text{ times}}$ by $c_k$. Notice that if $\ell>k$, then $c_\ell=10^{\ell-k}\cdot c_k+c_{{\ell-k}}$. Thus, $\gcd(c_\ell,c_k)=\gcd(c_{\ell-k},c_k)$. Moreover, $k\mid\ell$ clearly implies $c_k\mid c_\ell$. These observations indicate that in general $\gcd(c_\ell,c_k)=c_{\gcd(\ell,k)}$. In particular, if $\ell$ and $k$ are coprime, the so are $c_\ell$ and $c_k$. Now suppose there is a partition $A\sqcup B$ of $\{1,\dots,n\}$ for which
$$\prod_{a\in A}7^ac_a=\prod_{b\in B}7^bc_b.$$
Aiming for a contradiction, notice that by Chebyshev's theorem there exists an odd prime $p$ with $\frac{n}{2}<p<n$. For any $k\in\{1,\dots,n\}\setminus\{p\}$ one has $\gcd(c_p,c_k)=c_{\gcd(p,k)}=1$. WLOG, suppose $p\in A$. Hence $c_p\mid \prod_{b\in B}7^bc_b$. But $c_p$ is coprime with respect to any $c_b$ appearing on the RHS. Hence $c_p$ must divide $\prod_{b\in B}7^b$; in particular, it should be a power of $7$. But the same can be said about any other prime number $q$ between $\frac{n}{2}$ and $n$. There indeed exist two distinct primes $p$ and $q$ in the interval $\left(\frac{n}{2},n\right)$ for moderately large values of $n$ (e.g. $n\geq 40$, see here). They are indeed coprime and so are $c_p$ and $c_q$. Thus the last two numbers cannot be powers of $7$ simultaneously. For small values of $n$ for which there exists precisely one prime $p$ satisfying $\frac{n}{2}<p<n$, one should directly check that $c_p$ is not a power of $7$.
Update Following the comments by fedja and Aleksei Kulikov, the solution can be improved to show that $\prod_{k=1}^n7^kc_k$ is not a perfect square for $n\geq 6$. Assuming the contrary, $\prod_{k=1}^n\frac{c_k}{7^{e_k}}$ should be a perfect square where $e_k$ denotes the exponent of $7$ in the prime factorization of $c_k$. With the prime number $p$ as before, $\frac{c_p}{7^{e_p}}$ is coprime with respect to all other numbers appearing in the product. Hence $\frac{c_p}{7^{e_p}}$ should be a square itself. But that implies $c_p$ is of the form $x^2$ or $7x^2$. The former is impossible since $c_p\equiv 3 \,(\textrm{mod}\ 4)$. As for the latter, $7\nmid c_p$ because $7\mid c_6=111111$ and$\gcd(c_p,c_6)=c_{\gcd(p,6)}=c_1=1$.
