Geometric interpretation for sum of fourth powers

Summing the first $n$ first powers of natural numbers:

and there is a geometric proof involving two copies of a 2D representation of $(1+2+\cdots+n)$ that form a $n\times(n+1)$-rectangle.

Similarly,

has a geometric proof (scroll down just a bit til you see the wooden blocks) involving six copies of a 3D representation of $(1^2+2^2+\cdots+n^2)$ that form a $n\times(n+1)\times(2n+1)$-rectangular solid.

And similarly,

has a proof involving four copies of a 4D representation of $(1^3+2^3+\cdots+n^3)$ that form a $n\times n\times(n+1)\times(n+1)$-rectangular hypersolid in four-space. I can’t find a resource to demonstrate this last one better, but I’ve sketched it out for $n=3$, sketching the four dimensions using a $4\times4$ grid for two dimensions, and within each cell a $3\times 3$ grid for the other two. (Try it – it’s fun!) Here’s a smaller $n=2$ version. Note that $1^3+2^3$ is represented by some 4D blocks $1\times1^3+1\times2^3$ that are configured in a way that makes use of all four dimensions.

This is a $2\times2\times3\times3$ rectangular hypersolid made up of four copies of $1\cdot1^3+1\cdot2^3$.

So we move on to fourth powers:

This time the polynomial does not completely factor. I would like to know if anyone can find a similar geometric proof. It would involve 30 copies of a 5D representation of $(1^4+2^4+\cdots+n^4)$ forming a 5D hypersolid, one of whose 2-faces somehow works out to have area $3n^2+3n-1$, with the orthogonal edge lengths being $n$, $n+1$, and $2n+1$. I know it seems hopeless…30 copies? $3n^2+3n-1$? But it would make for some nice art.

It would be a nice start if there were some sort of connection between $30$ and a symmetry group of $\mathbb{R}^5$. For instance, if there were a subgroup of $\operatorname{SO}_5$ whose order was some large divisor of $30$, then that might help place the $30$ blocks of volume $1^5$ into their configuration.