Challenge

We have to solve \(c = m^e \ (mod \ p)\) where e = \(e_1^{e_2^{e_3}}\)
p-1 is smooth which allows phi(p-1) to be calculated. I used this approach https://math.stackexchange.com/questions/81228/modular-exponentiation-by-hand-ab-bmod-c

from Crypto.Util.number import *

p = 127557933868274766492781168166651795645253551106939814103375361345423596703884421796150924794852741931334746816404778765897684777811408386179315837751682393250322682273488477810275794941270780027115435485813413822503016999058941190903932883823
e1 = 219560036291700924162367491740680392841
e2 = 325829142086458078752836113369745585569
e3 = 237262361171684477270779152881433264701
c = 962976093858853504877937799237367527464560456536071770645193845048591657714868645727169308285896910567283470660044952959089092802768837038911347652160892917850466319249036343642773207046774240176141525105555149800395040339351956120433647613

e = pow(e1, pow(e2, e3, euler_phi(p-1)), p-1)
m = pow(c,pow(e,-1,p-1),p)
print(long_to_bytes(int(m)))
#crew{7hi5_1s_4_5ma11er_numb3r_7han_7h3_Gr4ham_numb3r}