Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$$\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1).$$

Because$$
a^{2m} - 1 = (a^2)^m - 1 = (a^2 - 1)((a^2)^{m - 1} + (a^2)^{m - 2} + \cdots + 1).
$$
