Knowing that the last digit of $7^2$ is $9$, we don’t actually need to compute $7^3$ to find its last digit. Look for a pattern in the last digits of powers of $7$.

It often helps to tackle a smaller problem first to get an idea, so let’s look at $7^2$, $7^3$, $7^4$ etc. and see what we can work out.

$7^2 = 49$, so the last digit is $9$, or in other words \[7^2 \equiv 9 \pmod{10}.\] Now if we want to work out $7^3 \bmod 10$, we don’t actually need to calculate $7^3$; we can just find $9\times7$. \begin{align*} 7^3 &= 7^2 \times 7\\ &\equiv 9 \times 7 \pmod{10}\\ &\equiv 63 \pmod{10}\\ &\equiv 3 \pmod{10}. \end{align*} \begin{align*} 7^4 &= 7^3 \times 7\\ &\equiv 3 \times7 \pmod{10}\\ &\equiv 21 \pmod{10}\\ &\equiv 1 \pmod{10}. \end{align*} \begin{align*} 7^5 &= 7^4 \times 7\\ &\equiv 1 \times7 \pmod{10}\\ &\equiv 7 \pmod{10}. \end{align*} The last digit of each power of $7$ depends only on the last digit of the previous power, so once we get a repeat, the pattern keeps going forever. \begin{align*} 7^0 \bmod 10 &= 1\\ 7^1 \bmod 10 &= 7\\ 7^2 \bmod 10 &= 9\\ 7^3 \bmod 10 &= 3\\ 7^4 \bmod 10 &= 1\\ 7^5 \bmod 10 &= 7\\ 7^6 \bmod 10 &= 9\\ 7^7 \bmod 10 &= 3\\ \cdots& \end{align*} Now we have a repeating pattern with $4$ different digits. Since $50 \bmod 4 = 2$, $7^{50}$ will have the same last digit as $7^2$. Hence the last digit of $7^{50}$ is $9$.