1Tabulate $2^n \bmod 100$ from $n = 2$: $4, 8, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52$ for $n = 2, \dots, 21$.
2At $n = 22$ the value is $04$ again, matching $n = 2$, so $2^n \bmod 100$ repeats with period $20$ for $n \ge 2$.
3Reduce the exponent within the cycle: $2026 = 2 + (2024)$ and $2024 \equiv 4 \pmod{20}$, so $2^{2026} \equiv 2^{2+4} = 2^6 \pmod{100}$.
4From the table $2^6 = 64$, so the last two digits are $64$.