//1530 x的n次方 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; long long use(int x,int n){ int s=0; if(n==0){ return 1; } n-=1; return use(x,n)*x; } int main(){ int n,m; cin>>n>>m; cout<<use(n,m); return 0; }