2的10000次方实现逻辑

package main

import (
    "fmt"
    "strconv"
)

func main() {
    var input string
    for {
        fmt.Println("")
        fmt.Println("请输入2的几次幂:")
        if _, err := fmt.Scanf("%s", &input); err != nil {
            fmt.Printf("%s\n", err)
            return
        }
        if input == "exit" {
            fmt.Println("退出成功~")
            return
        } else {
            ret := calculate( input )
            fmt.Println("结果为:")
            for i := len(ret)-1;i>=0 ;i--  {
                fmt.Print(ret[i])
            }
            fmt.Println()

        }
        fmt.Println("输入【exit】程序退出~")
    }
}

func calculate(input string) []int {
    ret :=[]int{1}
    power,_ := strconv.Atoi(input)
    if power<=0{
        return ret
    }
    carry :=0
    for i:=0;i< power ;i++  {
        for j:=0;j<len(ret);j++  {
            tmp := (2 * ret[j]) + carry
            if tmp>=10 {
                ret[j] = tmp%10
                carry = 1
            }else{
                ret[j] = tmp
                carry = 0
            }
        }
        if carry>0 {
            ret = append(ret,carry)
            carry=0
        }
    }

    return ret
}