递归法：

#include<stdio.h>
int mi(int n, int k)
{if (k == 1)   //如果是1次方{return n;  //返回n的1次方（也就是n）}else{return n * mi(n, k - 1);   //n*n的k-1次方}
}
int main()
{int n = 0, k = 0, sum = 0;printf("请输入n值和k值\n");scanf("%d%d", &n, &k);sum = mi(n, k);printf("%d的%d次方的值为%d\n", n, k, sum);return 0;
}

非递归法：

#include<stdio.h>
int mi(int n, int k)
{int i = 0, sum = 1;   //sum的初始值要为1，不能为0，不然0*任何数还是0for (i = 0; i < k; i++)    //循环k次{sum *= n;   //每循环一次就*n，}return sum;   //返回n的k次方的值
}
int main()
{int n = 0, k = 0, sum = 0;printf("请输入n值和k值\n");scanf("%d%d", &n, &k);sum = mi(n, k);printf("%d的%d次方的值为%d\n", n, k, sum);return 0;
}