$41$ divides $2^{20}-1$

Question

This is my first modular arithmetic proof so I'm checking to see if it works:

To begin, I want to show that this is true: \begin{align*} 2^{20}-1 \equiv 0 \pmod{41} \\ 2^{20} \equiv 1 \pmod{41} \end{align*}

Now, I note this congruence with the power of $2$ and then do the identity "if $a \equiv b \pmod n$ then $a^k \equiv b^k \pmod n$" \begin{align*} 2^5 \equiv -9 \pmod{41} \\ \left(2^5\right)^4 \equiv (-9)^4 \pmod{41} \end{align*}

Now, I note another congruence with the power of $-9$: \begin{align*} (-9)^2 \equiv -1 \pmod{41} \\ (-9)^4 \equiv 1 \pmod{41} \end{align*}

I can put these together because ``if $a \equiv b \pmod n$ and $b \equiv c \pmod n$, then $a \equiv c \pmod n$" \begin{align*} 2^{20} \equiv (-9)^4 \pmod{41} \\ \equiv 1 \pmod{41} \end{align*} And I get what I wanted to show that was true.

Although this is different than the book, I think this is still correct? At least, I think I'm following the basic properties correctly.