The standard definition of the factorial function $N!$ is a closed form whose time complexity is $\mathcal{O}(N!) = N$. Intuitively, it would seem that this could not be improved.

However, how can we be sure? What if there is some clever equivalent definition of the function where $\mathcal{O}(N!) = \log(N)$, or $\log(\log(N))$, or $N^{0.998}$? If so, how could we know it is possible, or find an example? If not, why not?

For clarity, I'm not talking about functions whose big-O is $N!$, I'm talking about the big-O of $N!$ itself.

Edit: Yes I know I'm abusing big O notation, but writing "$\mathcal{O}(f)$ where $f(N) = N!$" wouldn't fit easily in a title and $\mathcal{O}(!)$ looks silly and confusing at first glance.

The standard algorithm to compute the factorial function $N!$ via repeated multiplications has complexity $\mathcal{O}(N)$, in the model in which each operation costs 1, no matter how many digits the operands have. Intuitively, it would seem that this cost could not be improved.

However, how can we be sure? What if there is some clever equivalent algorithm with complexity $\mathcal{O}(\log(N))$, or $\mathcal{O}(N^{0.998})$? If so, how could we know it is possible, or find an example? If not, why not?

The standard definition of the factorial function $N!$ is a closed form whose time complexity is $\mathcal{O}(N!) = N$. Intuitively, it would seem that this could not be improved.

However, how can we be sure? What if there is some clever equivalent definition of the function where $\mathcal{O}(N!) = \log(N)$, or $\log(\log(N))$, or $N^{0.998}$? If so, how could we know it is possible, or find an example? If not, why not?

For clarity, I'm not talking about functions whose big-O is $N!$, I'm talking about the big-O of $N!$ itself.

The standard definition of the factorial function $N!$ is a closed form whose time complexity is $\mathcal{O}(N!) = N$. Intuitively, it would seem that this could not be improved.

However, how can we be sure? What if there is some clever equivalent definition of the function where $\mathcal{O}(N!) = \log(N)$, or $\log(\log(N))$, or $N^{0.998}$? If so, how could we know it is possible, or find an example? If not, why not?

For clarity, I'm not talking about functions whose big-O is $N!$, I'm talking about the big-O of $N!$ itself.

Edit: Yes I know I'm abusing big O notation, but writing "$\mathcal{O}(f)$ where $f(N) = N!$" wouldn't fit easily in a title and $\mathcal{O}(!)$ looks silly and confusing at first glance.