Hello. We will prove that the maximum value of the ratio $\frac{n}{m}$ is $2$ and is achived only for $n=2,m=1$. First of all,it's easy to verify that the triples $(m,n,k)=(1,2,k)$ satisfy the inequality,for all possible values of $k$. Now,consider a triple $(m,n,k)$ that also satisfies,such that $n\geq 2m$,and suppose that $\left |m^k-n!\right |=n-l \ , \ l\in \{1,2,\ldots ,n-1\}$.There are two possible cases: $\bullet$ Suppose that $m^k-n!=n-l$.Then $m^k=(n-l)\cdot \left(\frac{n!}{n-l}+1\right)$. The relation $n\geq 2m$ implies that $n!$ "contains" at least two multiples of $m$.Thus $m\mid \frac{n!}{n-l}$.Suppose that $\frac{n!}{n-l}=am$. However,from the equality $m^k=(n-l)\cdot \left(\frac{n!}{n-l}+1\right)=(n-l)(am+1)$ we obtain $am+1\mid m^k$. Take a common prime factor $p$ of $am+1,m^k$.Then $p\mid m\Rightarrow p\mid am\Rightarrow p\mid 1$,absurd. Since $am+1> 1$,we reach a contradiction. $\bullet$ Suppose that $m^k-n!=l-n$.Then $m^k=(n-l)\cdot \left(\frac{n!}{n-l}-1\right)$. In a similar way as above,we obtain $m\mid \frac{n!}{n-l}$. Suppose that $\frac{n!}{n-l}=am$.From the relation $m^k=(n-l)\cdot \left(\frac{n!}{n-l}-1\right)=(n-l)(am-1)$, we deduce that $am-1\mid m^k$.Hence,either $am-1=1$ or there exists a prime $p$ which divides both $am-1,m^k$. The latter is rejected in a similar way as in the first case. Thus,$am-1=1\Rightarrow am=2\Rightarrow n!=2(n-l)$,which holds only for $(n,l)=(2,1),(3,0)$. However,for $n=3$ there are no acceptable triples$(m,n,k)$ with $n\geq 2m$ and, for $n=2$,those triples are all of the form $(m,n,k)=(1,2,k)$ and only them. Thus,the maximum value of the ratio $\frac{n}{m}$ is indeed equal to $2$ and is achieved only for $m=1,n=2$.