Show that $11^{10} \equiv 1 (\mod 100)$ $11^{10} \equiv 1 \pmod{100}$
I tried to solve by using euler's theorem, But I got stuck.
$\gcd(11, 100) = 1$
$11^{φ(100)} \equiv 1 \pmod{100}$
$11^{40} \equiv 1 \pmod{100}$
I don't know how to go on as $11^{40}$ is bigger than $11^{10}$

$$\qquad{11^2\equiv 121\equiv 21\mod 100\\\\
(11^2)\times 11\equiv 231\equiv 31\mod 100\\
(11^4)\equiv 31\times 11\equiv 341\equiv 41\mod 100\\
(11^5)\equiv 41\times 11\equiv 451\equiv 51\mod 100\\
(11^6)\equiv 51\times 11\equiv 61\mod 100\\
(11^7)\equiv 61\times 11\equiv 71\mod 100\\
(11^8)\equiv 71\times 11\equiv 81\mod 100\\
(11^9)\equiv 81\times 11\equiv 91\mod 100\\
(11^{10})\equiv 91\times 11\equiv 1001\equiv 1\mod 100\\}$$
