How is $9^{40}\equiv\ 1 \pmod {100}$? By Euler's theorem, $a^{\varphi(100)} \equiv\ 1 \pmod {100}$.
We know that the last two digits of $9^{40}$ are non-zero. So they can't even be $01$.
Since $1\equiv\ 1 \pmod {100}$, how come $9^{40}\equiv\  1 \pmod {100}$?
I have looked at:
Find the last two digits of $9^{{9}^{9}}$,
Find the last two digits of the number $9^{9^9}$ ,
Find the last two digits of $9^{9^{9}}$

It is straightforward
$9^{40} = 81^{20} = 6561^{10} \equiv  61^{10} \pmod{100} = 3721^5 \pmod{100} \equiv\  21^5 \pmod{100}$ 
$= 441 \times 441 \times 21 \pmod{100}\equiv\ 41 \times 41 \times 21 \pmod{100}$
$= 35301 \pmod{100} \equiv\  1 \ \pmod{100}$
and this is just one of the many many routes you could take.
