Prove that cube cannot be partitioned into $n>1$ smaller distinct cubes. 
Prove that cube cannot be partitioned into $n>1$ cubes, such that each of them has different side length.

I believe tallhis is not hard problem, but I just do not have an idea how to start. I tried to prove it by contradiction. I assumed that large cube is tiled with a $n>1$ cubes such that at all of small cubes have different side length. Then I tried to use induction. 
For $n=2$ it is trivial that it is impossible because in that case one side of a small cube must be twice of other side. For $n=3$ it is impossible because wherever we place first small cube, the remaining area cannot be a cuboid so we cannot partition the remaining area into $2$ cubes. For $n=4$ I do not see an easy way to prove it. 
I do not believe that induction is the best method, but any solution will be appreciated.

As user vadim123 said in the comments, a full answer to this question appears on this page on Wikipedia.

[Given a cube $C$] suppose that there is such a dissection. Make a face of $C$ its horizontal base. The base is divided into a perfect squared rectangle $R$ by the cubes which rest on it. Each corner square of $R$ has a smaller adjacent edge square, and $R$'s smallest edge square is adjacent to smaller squares not on the edge. Therefore, the smallest square $s_1$ in $R$ is surrounded by larger, and therefore higher, cubes on all four sides. Hence the upper face of the cube on $s_1$ is divided into a perfect squared square by the cubes which rest on it. Let $s_2$ be the smallest square in this dissection. The sequence of squares $s_1, s_2, \dotsc\;$ is infinite and the corresponding cubes are infinite in number. This contradicts our original supposition.
If a $4$-dimensional hypercube could be perfectly hypercubed then its 'faces' would be perfect cubed cubes; this is impossible. Similarly, there is no solution for all cubes of higher dimensions.

