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 $9^{9^{9}}$
 A: $9^2\equiv\ -19\ (mod100)$
$9^4 \equiv\ 61\ (mod 100)$
$9^8\equiv\  21 \ (mod 100)$
$9^{10}\equiv\ 1 \ (mod 100)$
$9^{40}\equiv\ 1 \ (mod 100)$
A: Calculate $$9^{10} = 3486784401$$
So
$$9^{40} \mod 100 \equiv (9^{10}\mod 100)^4 \mod 100 $$
$$\equiv(1)^4 \mod 100 \equiv 1 \mod 100$$ 
A: By binomial theorem:
$$(10-1)^{40} = \sum_{i=0}^{40} \binom{40}{i}(-1)^{40-i}10^{i}$$
Modulo $10^2$, you only need to look at $i=0,1$.
So:
$$9^{40}\equiv (-1)^{40} + \binom{40}{1}(-1)^{39}\cdot 10 \equiv (-1)^{40}\equiv 1\pmod{100}$$
Note that this works for $9^{10}$, too.
A: 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.
A: Note that $$99^2=9801\equiv 1\pmod{100}\implies99^{40}\equiv 1\pmod{100}\implies9^{40}11^{40}\equiv 1\pmod{100}$$
$$9^{40}\equiv 1\pmod{100}$$
A: If you want to avoid the equation
$9^{40}=147808829414345923316083210206383297601$,
the question is how low you prefer your numbers to be.
One alternative is using $40=4*2*5$ (and 7 digits for the last operation).
$9^4=61$  (mod 100)
$61^2=21$  (mod 100), and
$21^5=01$  (mod 100)
