Just 2 digits

Let the number given be shortened to $19^{29^n}$ . Then

$\begin{aligned} 19^{29^n} & \equiv 19^{\color{#3D99F6}29^n \bmod \lambda (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }\gcd{19, 100}=1 \text{, Euler's theorem applies.} \\ & \equiv 19^{\color{#3D99F6}29^n \bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{Carmichael's lambda function }\lambda (100) = 20 \\ & \equiv 19^{9^{\color{#3D99F6}n} \bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{Note that }n\text{ is odd and let }n=2k+1 \\ & \equiv 19^{81^k \times 9 \bmod 20} \text{ (mod 100)} \\ & \equiv 19^{(80+1)^k \times 9 \bmod 20} \text{ (mod 100)} \\ & \equiv 19^9 \text{ (mod 100)} \\ & \equiv (20-1)^5(20-1)^4 \text{ (mod 100)} \\ & \equiv (-1)^5(-80+1) \text{ (mod 100)} \\ & \equiv \boxed{79} \text{ (mod 100)} \end{aligned}$

References:

• Euler's theorem
• Carmichael's lambda function