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Nice problem! The numbers work out very well...
We compute 7 7 ≡ 7 3 ≡ 4 3 ( m o d 1 0 0 ) . Now 7 7 7 ≡ 7 4 3 ≡ 7 3 ≡ 4 3 ( m o d 1 0 0 ) . In the exponent we can work modulo 2 0 = λ ( 1 0 0 ) , using the Carmichael lambda.
The sequence becomes stationary after one step, at 4 3