I’ll present the official solution. Let $ S$ be the family of all configurations of $ n$ cubes (cells), no two of them lying on the same layer. Let's count how many elements $ S$ has. We can choose the firs cell $ c_1$ in $ n^3$ ways. After removing all layers $ c_1$ takes part in there remain $ (n-1)^3$ cubes. So the next cell $ c_2$ can be chosen in $ (n-1)^3$ ways and so on. This way, every configuration in $ S$ is counted $ n!$ times, hence $$ \displaystyle |S|=\frac{n^3\cdot (n-1)^3\cdots 2^3\cdot 1^3}{n!}=(n!)^2.$$Denote by $ B$ the family of "bad" configurations of $ n$ cells, no two of them in the same layer, in which there exists two identically colored cells. We can choose two cells of the same color by first choosing the color and then all possible pairs among the four cells colored in this color. Thus, we can choose two identically colored cells in $ \displaystyle \frac{n^3}{4}\cdot \binom{4}{2}=3\frac{n^3}{2}$ ways. Therefore $$ \displaystyle |B|\le 3\frac{n^3}{2} \big((n-2)!\big)^2 .$$Note that this inequality may be strict. For example it may happen a bad configuration consists of two pairs of identically colored cells. Then we count this configuration twice. Further, it's enough to show $ |B|<|S|$ because then $ S\setminus B$ could not be empty. Thus, it's enough to prove that for $ n\ge 4$ it holds $$ \displaystyle 3\frac{n^3}{2} \big((n-2)!\big)^2 < (n!)^2.$$It is equivalent to $ \displaystyle 3\frac{n}{2} < (n-1)^2 $ and it can be easily checked that it holds for $ n\ge 4.$ Remark. Some more comments and a generalization of this problem can be found in my blog.