Let $n$ and $m$ be positive integers with $m \le n! := n(n-1)\ldots(2)(1)$.
Question 1: What's a good lower bound for $n$ as a function of $m$ ?
For example, it's easy to see that $$ \begin{aligned} \log m\le \log n! = \sum_{1 \le k \le n} \log k \overset{\text{J.I.}}{\le} \left(\sum_{1\le k\le n}1\right)\log\left(\frac{\sum_{1\le k\le n}k}{\sum_{1\le k \le n}1}\right) = n \log\left( \frac{n + 1}{2}\right) \ll n^2, \end{aligned} $$ and so we must have $n \gg (\log m)^{1/2}$. Can one do much better than this ?
Question 2: Define $\log^1 m = \log m$, $\log^2 m = \log \log m$, $\log^3 m = \log \log \log m$, etc. Does there exist an integer $k \ge 2$ and real numbers $a, u, v$ with $a, u > 0$, all independent of $m$, such that $$n \ge h_{a, u, k, v}(m) := a + u \log m/ (\log^k m)^v$$ for sufficiently large $m$ ?
N.B.: Ideally, we'd want $a$, $u$, and $k$ to be "big", and $v$ to be "small".
Update
The accepted answer solves the 2nd problem (and therefore both problems!) by showing that
If $e^e < m \le n!$, then $n \ge h_{0, 1, 2, 1}(m) = \log m/\log^2 m$.
Claim: if $n!\ge m > e^e$, then $n \ge \displaystyle\frac{\log m}{\log\log m}$.
Proof: We prove the contrapositive: Suppose that $m>e^e$ and $n < \frac{\log m}{\log\log m}$. Then $$ n! \le n^n < \bigg( \frac{\log m}{\log\log m} \bigg)^{(\log m)/\log\log m} < (\log m)^{(\log m)/\log\log m} = m. $$
This lower bound is essentially best possible, in that for any $\varepsilon>0$, there are integers $m,n$ satisfying both $n!\ge m$ and $n < (1+\varepsilon) \frac{\log m}{\log\log m}$. One could get a second-order term with a little more work.