We just finished our Time constructability lesson in class last week, and we, for example's sake, showed that $n^k, 2^n$ are fully time constructible, i.e. there exists a (multi-tape deterministic) Turing machine that for $n$ given, halts after exactly $f(n)$ steps, and just asked if we could now prove that $n!$ is fully time constructible (and moved on).I am not sure how the proof goes, but I am thinking it has to use $n^k$ time constructability to some extent, or some identity involving factorials, since we showed $n^k$ is (fully) time constructible using $n^k = n + \sum_{i=1}^{i=k-1}(n - 1)n^i$.Hints would be appreciated too, really. Thanks in advance. 
Prove n! is fully time constructible
complexity theory;turing machines;time complexity