An alternative approach I much prefer (as it's easier to grasp) but that's a bit longer:
$ 3^{50} = (3^2)^{25} = 9^{25} \equiv 2^{25} $ mod $7$
Just keep removing squares from the even exponents, and then computing a smaller equivalent mod $7$:
$ 2^{25} = 2 \cdot 2^{24} = 2 \cdot 4^{12} = 2 \cdot 16^6 \equiv 2 \cdot 2^{6} $ mod $7$
Nearly there.. When the exponent is odd, take out 1 factor to make it even again
$ 2 \cdot 2^{6} = 2 \cdot 4^{3} = 2 \cdot 4 \cdot 4^2 = 8 \cdot 16 \equiv 1 \cdot 2 \equiv 2 $ mod $7$
Done.