Prove n! is fully time constructible
Let us assume we have find $n!$ on input $n$. Our complexity is terms of length of the input ( we assume it to be $L=O(logn)$ ). Multiplication of a $k$ bit by $l$ bit number via standard multiplication takes $O(kl)$ operations ( also after multiplication number of bits in the resultant if $O(k+l)$ ) . We multiply linearly in a loop from $1$ to $n$ to get $n!$. So number of operations carried out are upper bounded by $\# = L^2  ( 1 + 2 ...(n-1) ) = \frac{L^2n(n-1)}{2} = O(L^22^{O(L)})=o(L!)$. Thus it is space and time constructible.