Observe that
7¹ ≡ 7 (mod 100)
7² ≡ 49 (mod 100)
7³ ≡ 343 ≡ 43 (mod 100)
and (using Euclid's algorithm)
7 ⁻¹ ≡ 43 (mod 100)
so that
7⁴ ≡ 7³ × 7¹ ≡ 43 × 7¹ ≡ 1 (mod 100)
This means
7⁷ ≡ 7⁴ × 7³ ≡ 1 × 43 ≡ 43 (mod 100)
Next,
(7⁷)⁷ ≡ 43⁷ ≡ (7 ⁻¹)⁷ ≡ (7⁷) ⁻¹ ≡ 43⁻¹ ≡ 7 (mod 100)
(3 7's)
Next,
((7⁷)⁷)⁷ ≡ 7⁷ ≡ 43 (mod 100)
(4 7's)
A pattern emerges: a power tower involving an even number of 7's will reduce to 43 (mod 100), so the last two digits of ((7⁷)⁷ ... )⁷ with 2020 7's will have 43 as its last two digits.