divisability

The post by Arnab can be fixed (one can prove the inequality)

Nonsense in the sense Hermite's identity does absolutely nothing

:D

Here is a proof by combinatorial argument

A := \left \{t : t \; \; \; \text{is a disjoint partition of} \left \{1, \ldots , k^2 \right \} \text{with each classes containing}\; \; \; k\; \; \; \text{elements of} \left \{1, \ldots , k^2 \right \} \right \}

B := \left \{t : t \; \; \; \text{is a disjoint partition of} \left \{1, \ldots , k^3 \right \} \text{with each classes containing}\; \; \; k\; \; \; \text{elements of} \left \{1, \ldots , k^3 \right \} \right \}

Then

|A| = \frac{k^2 !}{ k ! \cdot (k!)^{k}}

|B| = \frac{k^3 !}{ k^2 ! \cdot (k!)^{k^2}}

And
|A| |B| = \frac{k^3 !}{(k!)^{k^2 + k+ 1}}
so we are done.

Let p\le k be a prime....

Proving \lfloor \frac{k^3}{p}\rfloor+\lfloor \frac{k^3}{p^2}\rfloor+\lfloor \frac{k^3}{p^3}\rfloor+\cdots\ge(k^2+k+1)\left(\lfloor \frac{k}{p}\rfloor+\lfloor \frac{k}{p^2}\rfloor+\lfloor \frac{k}{p^3}\rfloor+\cdots\right)
will be sufficient.

Hermite's Identity says that \lfloor nx\rfloor=\lfloor x+\frac1n\rfloor+\lfloor x+\frac2n\rfloor+ \lfloor x+\frac3n\rfloor+ \cdots \lfloor x+\frac{n-1}{n}\rfloor
where n is a positive integer.

Now break with the help of the idenity that:
\sum_{r=1}^{\infty}\lfloor k^2\frac{k}{p^r}\rfloor =\sum_{r=1}^{\infty}\lfloor\frac{k}{p^r}+\frac{1}{k^2}\rfloor+\cdots+\sum_{r=1}^{\infty}\lfloor\frac{k}{p^r}+\frac{1}{k}\rfloor+\sum_{r=1}^{\infty}\lfloor\frac{k}{p^r}+\frac{1}{k}+\frac{1}{k^2}\rfloor+\cdots+\sum_{r=1}^{\infty}\lfloor\frac{k}{p^r}+1-\frac{1}{k^2}\rfloor

Let v_p(N) denote the highest power of p that divides N

for p such that $p \not |\ k$ it obvious that

v_p (k^3!) = v_p((k^3-1)!)=v_p([(k-1)(k^2+k+1)]!)

\ge (k^2+k+1)v_p((k-1)!) = (k^2+k+1)v_p(k!)

The proof needs to be devised for p|k

@man111: I don't think this has an easier proof by induction.

Do you have such a proof ?