Find the least natural number such that its cube is less than its factorial. Basically, find a natural number $k$ such that $k^3 > k!$ but $(k + 1)^3 < (k + 1)!$.
Now I know that the answer is 5.
The issue is in the deriving this through algebra. Here's what I did:
$(k + 1)^3 < (k + 1)!$ $\Rightarrow$ $(k + 1)^3 < k!(k + 1)$ $< k^3(k + 1)$, since $k^3 > k!$, and $(k + 1) > 0$.
Thus, $(k + 1)^3 < k^3(k + 1)$ giving  $(k + 1)^2 < k^3$ because again, $(k + 1) > 0$.
Now the issue is, $k = 6$, for example, satisfies this equation even though its cube is less that its factorial, contrary to the condition this equation was derived using.
I feel I am making some error that is gonna make me pull my hair out once I realise it but I cannot figure out what it is.

First of all, the answer is so low there is no need to do general algebra and inequality manipulation to solve it. We see that $5^3>5!$ and $6^3<6!$, and that's it.
Second, your logic goes the opposite way of what you seem to think. You start with the assumption that $k!<k^3$ and $(k+1)!>(k+1)^3$, and from that derive that $(k+1)^2<k^3$. And this is entirely true. Any $k$ that satisfies $k!<k^3$ and $(k+1)!>(k+1)^3$ will also satisfy $(k+1)^2<k^3$.
However, there are also many other $k$ that satisfy $(k+1)^2<k^3$. This is not a problem. It just means your algebra and inequality manipulation haven't narrowed the set of possible solutions all the way from $\Bbb N$ to a single number. Or, perhaps more correctly, your manipulations have introduced additional solutions.
