Find the super palindromes! Consider the number 99999999. That number is obviously a palindrome.  The largest prime factor of 99999999 is 137. If you divide 99999999 by 137, you get 729927. This number is also a palindrome.
The largest prime factor of 729927 is 101. 729927/101=7227 which again is a palindrome.
The largest prime factor of 7227 is 73. 7227/73=99 which again is a palindrome.
By further dividing by the largest prime factor, you get 9, 3 and finally 1, which, being one-digit numbers, are also palindromes. Since 1 has no prime factors, the procedure ends here.
Now generalizing this observation, I define a super-palindrome as a palindrome which is either 1, or which gives another super-palindrome if divided by its largest prime factor.

Credits: https://math.stackexchange.com/questions/200835/are-there-infinitely-many-super-palindromes
Given a number N, determine if it is a super palindrome or not, and print a truthy or falsey value accordingly.
Your program should print a truthy value for these inputs:
1
101
121
282
313
353
373
393
474
737
919
959
1331
1441
2882
6446
7887
8668
9559
9779

Your program should print a falsey value for these inputs:
323
432
555
583
585
646
642
696
777
969
989
2112
3553
4554
5242
5225
5445
8080
8118
9988

Remember, this is code-golf, so the code with the shortest amount of bytes wins.
Jelly, 13 12 9 8 bytes

Æf×\D⁼U$


Try it online! or verify all test cases.

How it works

Æf×\D⁼U$  Main link. Argument: n

Æf        Yield all prime factors of n, with multiplicities and in ascending order.
  ×\      Take the cumulative product.
    D     Decimal; convert each product into the array of its base 10 digits.
       $  Combine the two links to the left into a monadic chain.
      U     Upend; reverse all arrays of decimal digits.
     ⁼      Test for equality.