Too many 63's

Let the number given be $N$ . We need to find $N \bmod 10000$ . Since $\gcd(63,2,5)=1$ , we can use Euler's theorem and Carmichael lambda function . The required Carmichael lambda values $\lambda (10000) = 500$ , $\lambda (500) = 100$ , $\lambda (100) = 20$ , $\lambda (20) = 4$ , $\lambda (4) = 2$ and $\lambda (2) = 1$ . Then, we have:

$\Large \begin{aligned} N & = 63^{\color{#3D99F6}63^{\color{#D61F06}63^{\color{#3D99F6}63^{\color{#D61F06}63^{\color{#3D99F6}63^{\color{#D61F06}63^{\color{#3D99F6}63^{\color{#D61F06}63}} \bmod 1} \bmod 2} \bmod 4} \bmod 20} \bmod 100} \bmod 500} \text{ (mod 10000)} \end{aligned}$

Starting from $63 \equiv 0 \text{ (mod 1)}$ . Then $63^0 \equiv 1 \text{ (mod 2)}$ , $63^1 \equiv 3 \text{ (mod 4)}$ , $63^3 \equiv (60+3)^3 \equiv 7 \text{ (mod 20)}$ ,

$\begin{aligned} 63^7 & \equiv (60+3)^7 \equiv 20\times 3^6+3^7 \equiv 9^3(23) \equiv (10-1)^3(23) \equiv 29(23) \equiv 67 \text{ (mod 100)} \end{aligned}$

$\begin{aligned} 63^{67} & \equiv (50-1)^{30}(50-1)^37(10-1)^{50}(10-1)^{15}9^2 \equiv 149(7)149(81) \equiv 467 \text{ (mod 500)} \end{aligned}$

$\begin{aligned} 63^{467} & \equiv (50-1)^{200}(50-1)^{32}7^3(10-1)^{400}(10-1)^{60}(10-1)^7 \text{ (mod 10000)} \\ & \equiv (1)(-1600+1)(343)(-4000+1)(7000-600+1)(35000-2100+70-1) \text{ (mod 10000)} \\ & \equiv (8401)(343)(6001)(6401)(2969) \text{ (mod 10000)} \\ & \equiv (1543)(2401)(2969) \text{ (mod 10000)} \\ & \equiv (4743)(2969) \\ & = \boxed{1967} \end{aligned}$