The Secret 7

If you write the last digit of the first powers of seven:

$7^0 = 1$ ; last digit: 1 ; 0 mod 4 = 0

$7^1 = 7$ ; last digit: 7 ; 1 mod 4 = 1

$7^2 = 49$ ; last digit: 9 ; 2 mod 4 = 2

$7^3 = 343$ ; last digit: 3 ; 3 mod 4 = 3

$7^4 = 2401$ ; last digit: 1 ; 4 mod 4 = 0

$7^5 = 16807$ ; last digit: 7 ; 5 mod 4 = 1

....

There is a period of 4, so you only need 100 mod 4 = 0, so the last digit is 1.

Adding 5, the last digit of $7^{100}+5$ is 6

$7^{100}+5$ $=7^{0} \times 7^{100}+5$ $=7^{0} \times (7^{4})^{25}+5$ $=7^{0} \times 2401^{25}+5$ $\equiv 7^{0} \times 1^{25}+5$ $\equiv 1 \times 1+5$ $\equiv 1+5$ $\equiv 6 \pmod{10}$ Hence, the last digit is $\large \color{#624F41}{\boxed{\color{#D61F06}{6}}}$