User input manipulation - python-3.9

Question

Please I'm in need of a lift right now. I want the user to be able to have the option of re-entering his numerical input if perhaps, he/she entered an alphabet by mistake.
Right now it just loops infinitely on the "Please select a valid option" and "Try again" outputs.
print("\n-----Factorial Calculator------\n")
def option():
opt = (input('Try again? y/n: '))
if option == 'y':
facto(num)
elif option == 'n':
exit()
else:
print("Please select a valid option")
option()
def facto(factorial):
fact = 1
for nums in range(1, num+1):
fact = fact * nums
return fact
try:
num = int(input("Number(Just input the number): "))
except ValueError:
print("Invalid input")
option()
print(f" {num}! = {facto(num)}")

Answer 1

First, rather than calling option recursively, simply put a loop inside it.
Second, when the user enters y into the "Try again?" input, it should not calculate the factorial there, but return the user back to the original input, by breaking out of the loop within the option call.
Putting these two together will fix option:
def option():
while True:
retry = input('Try again? y/n: ')
if retry == 'y':
break
elif retry == 'n':
exit()
else:
print("Please select a valid option")
Third, you need some kind of loop around your initial input, otherwise how will they enter the number again after indicating they want to try again?
Fix this by sticking the initial num assignment and option call in a while loop, with the same try; except statement you had before:
while True:
try:
num = int(input("Number(Just input the number): "))
break
except ValueError:
print("Invalid input")
option()
continue

Answer 2

I see someone else got an answer in ahead of me which identified the bug in option() and the bug which prevents user re-entering the number but I'll post this anyway because there are other ways your code can be restructured.
print("\n-----Factorial Calculator------\n")
# Each function should handle just one thing
# Eg, option() should just handle whether user wants to continue, not try to handle factorial calc
# Also it's best to let the mainline logic decide whether to exit() or not
# This function should just return 'y' or 'n'
# Note that the original code captured the user's option in variable `opt` but then checked variable `option`.
# That's why it looped endlessly
def option():
opt = ' '
while ( opt != 'y' and opt != 'n' ):
opt = (input('Try again? y/n: '))
if opt == 'y':
return opt
elif opt == 'n':
return opt
else:
print("Please select a valid option")
# You specified argument `factorial` in original code but then did not use it.
# It still worked because the variable `num` was defined at the top level in the mainline
# Functions should be self-contained and rely only on the arguments that are passed into them
def facto(n):
fact = 1
for nums in range(1, n+1):
fact = fact * nums
return fact
opt = 'y'
is_print = False
while ( opt == 'y' ):
try:
num = int(input("Number(Just input the number): "))
opt = 'n'
is_print = True
except ValueError:
print("Invalid input")
opt = option()
if ( is_print ):
print(f" {num}! = {facto(num)}")

Answer 3

Thanks everyone for the help, but after struggling with it for a couple of days and taking few days off due to illness, I somehow did it. Eureka!
def facto(n):
fact = 1
for i in range(1, n + 1):
fact = fact * i
return fact
def opt():
option = (input("Do you want to try again? y/n: "))
if option == 'y':
number()
elif option == 'n':
exit()
else:
print("\nPlease select a valid option.")
opt()
def number():
try:
num = int(input("Number: "))
print(f"{num}! = {facto(num)}")
except ValueError:
print("Invalid input! Please input integers only")
opt()
number()