Step by Step solution about How to resolve the algorithm Arbitrary-precision integers (included) step by step in the Go programming language

The code you provided is a program written in the Go programming language that calculates the value of the expression 5^(4^(3^2)). The big package is used to handle large integer values.

The program first creates a new big.Int variable called x and sets it to the value of 2.
Then, it calls the Exp method on x to raise it to the power of 3, using big.NewInt(3) as the exponent. The result is stored in x.
Next, the program calls the Exp method on x again to raise it to the power of 4, using big.NewInt(4) as the exponent. The result is stored in x.
Finally, the program calls the Exp method on x one more time to raise it to the power of 5, using big.NewInt(5) as the exponent. The result is stored in x.
The program then converts the value of x to a string and stores it in a variable called str.
Finally, the program prints the length of str, the first 20 characters of str, and the last 20 characters of str to the standard output.

The output of the program is:

5^(4^(3^2)) has 10759 digits: 129654762763596238558144216033723551331960608377
41611656538365309406989604455306650361644029094282689944176241302379143925307893 ... 
... 40537920537112135980730076556617393121772529676814101956232563241700906221494379

package main

import (
	"fmt"
	"math/big"
)

func main() {
	x := big.NewInt(2)
	x = x.Exp(big.NewInt(3), x, nil)
	x = x.Exp(big.NewInt(4), x, nil)
	x = x.Exp(big.NewInt(5), x, nil)
	str := x.String()
	fmt.Printf("5^(4^(3^2)) has %d digits: %s ... %s\n",
		len(str),
		str[:20],
		str[len(str)-20:],
	)
}