Multi-taped Turing machine are equivalent to the standard single-taped variety; that much I know and can prove to a level of rigor that satisfies me; so long as there are finitely many tapes, I can think of how to simulate a model like that on a standard Turing machine.
Now, let's say we have a model similar to that, but with infinitely many tapes. And, unlike in Yuval Filmus's answer to this question, I'm thinking of the case that is a direct "expansion" of the multi-tape case – meaning: there's a head for each tape, so also infinitely many heads.
I believe this model is not equivalent to Turing machines anymore, and that it is stronger. However, I have a hard time with thinking of how to prove this.
I wanted to approach this from a set cardinality angle: each TM can be encoded as some finite binary string (maybe after being reduced to some "canonical" form; an equivalent TM), so there are countably many turing machines: $\aleph_0$. At the same time, there are $2^{\aleph_0}$ functions $f: \{0,1\}^* \rightarrow \{0,1\}$. Based on this, it cannot be the case that each such $f$ has a Turing machine that computes it, so some (a lot, actually...) of such functions aren't computable.
I think the proposed infinite-tapes model could simulate any such function, using its abiilty to read & write infinite data at each step due to its infinite heads, which would mean that there are functions it can compute that the standard TM model cannot, and so they're not equivalent – that'd be great and suffice for my needs. However, I've only just learned the concept of a Turing machine and I'm having a hard time coming up with the exact "recipe" for how to compute an arbitrary given $f: \{0,1\}^* \rightarrow \{0,1\}$ with such a machine.
How might this be done? Alternatively: is there another neat proof that doesn't require more advanced knowledge?
