modular exponentiation $2^{103}\equiv 63\pmod {143}$ I am trying to prove that $2^{103}\equiv 63\pmod {143}$.One way is to ''break'' the power of $2$ but $103$ is prime.Is there any way or theorem?

One very common method for calculating large powers is repeated squaring.
To use it for your example here, we have:
$$
2 \equiv 2 \pmod{143} \\
2^2 \equiv 4 \pmod{143} \\
2^4 \equiv 16 \pmod{143} \\
2^8 \equiv 256 \equiv 113 \pmod{143} \\
2^{16} \equiv (-30)^2 \equiv 900 \equiv 42 \pmod{143} \\
2^{32} \equiv 42^2 \equiv 1764 \equiv 48 \pmod{143} \\
2^{64} \equiv 48^2 \equiv 2304 \equiv 16 \pmod{143} \\
$$
We therefore have:
$$
\begin{align}
2^{103} &\equiv 2^{64} \cdot 2^{32} \cdot  2^4 \cdot 2^2 \cdot 2 \pmod{143} \\
&\equiv (16 \cdot 48) \cdot 16 \cdot 4 \cdot 2 \pmod{143} \\
&\equiv (53 \cdot 16) \cdot (4 \cdot 2) \pmod{143} \\
&\equiv 133 \cdot 8 \pmod{143} \\
&\equiv 63 \pmod{143}
\end{align}
$$
