Shouldn't it be mod m?

For any prime $p\mid m$, and $1,2^n,\cdots,m^n$ is complete module $m$ for a large $n>1$, then there exists $i$, such that $(i^n, m)=p$, then $p || m$. So $m=p_1p_2\cdots p_t$, where $p_i$ are prime. We select $n=1+k(p_1-1)\cdots (p_t-1)$ will be OK

If $m$ is not square free,then just choose $k=p_{1}..p_{i}$,the product of all primes dividing $m$,but for large $n$ both $k,m$ is $0$ mod $m$.Hence,class isn't complete. For $m$ square free,choose $n=1+x(p_{1}-1)...(p_{n}-1)$.