Trouble with elementary number theory I'm trying to solve the following two problems:

Show $n^{12}\equiv 1\pmod{72}$ when $(72,n)=1$

and

Compute ${7^8}^9\pmod{100}$

For the first, I saw that $n^{24}=n^{\varphi(72)}\equiv1\pmod{72}$ by Euler's theorem, but this leaves us with $n^{12}n^{12}\equiv 1\pmod{72}$. This could mean $n^{12}\equiv 1\pmod{72}$ or $n^{12}\equiv 71\pmod{72}$. I'm not sure how to argue correctly to conclude it's only $n^{12}\equiv 1\pmod{72}$.
For the second one, I practically did brute force. ${7^8}^9\pmod{100}$ can be solved by considering $8^{9}\pmod{\varphi(100)=40}$. Since $8$ is cyclic with order $4: 8^5\equiv 8\pmod{40}$, we know $8^9=8^5\cdot8^4\equiv 8^5\equiv 8$. So look at $7^8\pmod{100}$. 7 is also cyclic with $7^4\equiv 1\pmod{100}$. So $7^8=(7^4)^2\equiv 1\pmod{100}$. I knew these "cyclic" things by manually computing and reducing, though.
Thank you for any help... (note this is not homework)

$(1)$ Using Carmichael function, $$\lambda(72)=\text{lcm}(\lambda(8),\lambda(9))=\text{lcm}(2,6)=6$$
$$\implies n^6\equiv1\pmod{72}\text{ if }(n,72)=1 $$
$(2)$ Observe that $\displaystyle7^2=49=50-1\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod{100}$
$\displaystyle\implies7^n\equiv1\pmod{100}$ if $4|n$
Now, $8^9=(2^3)^9=2^{27}$
