We need to find the remainder when $7^{103}$ is divided by $17$, i.e. compute $7^{103} \pmod{17}.$

Since $17$ is prime and $\gcd(7,17)=1$, we can use Fermat's Little Theorem: $7^{16} \equiv 1 \pmod{17}.$

Now reduce the exponent $103$ modulo $16$: $103 = 16\cdot 6 + 7.$ So, $7^{103} = 7^{16\cdot 6 + 7} = (7^{16})^6 \cdot 7^7.$

Using $7^{16} \equiv 1 \pmod{17}$, $(7^{16})^6 \cdot 7^7 \equiv 1^6 \cdot 7^7 = 7^7 \pmod{17}.$ So we only need to compute $7^7 \pmod{17}$.

Compute powers step-by-step: $7^2 = 49 \equiv 49-34 = 15 \pmod{17}.$ Then, $7^4 \equiv 15^2 = 225 \equiv 225-221 = 4 \pmod{17}.$ Now, $7^7 = 7^4\cdot 7^2\cdot 7 \equiv 4\cdot 15\cdot 7 \pmod{17}.$ First, $4\cdot 15 = 60 \equiv 60-51 = 9 \pmod{17}.$ Then, $9\cdot 7 = 63 \equiv 63-51 = 12 \pmod{17}.$ Thus, $7^7 \equiv 12 \pmod{17}.$

Therefore, $7^{103} \equiv 12 \pmod{17}.$ So the remainder is $\boxed{12}.$