Show $\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}$. Let $p$ be prime and $d \ge 2$. I want to show that
$$
 \frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}.
$$
I have a proof, but I think it is complicated, and the statement appears in a book as if it is very easy to see. So is there any easy argument to see it?
My proof uses
$$
 \frac{p^n - 1}{p-1} = 1 + p + \ldots + p^{n-1}.
$$
So
$$
 \frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} = \frac{(1 + p + \ldots + p^{d-1})(1 + p + \ldots + p^{d-2})}{p+1}
$$
If $d - 2$ is odd, set $u := (1 + p + \ldots + p^{d-1})$ and then proceed 
\begin{align*}
 & \frac{u(1 + p + \ldots + p^{d-2})}{p+1} \\
 & = \frac{u(1+p) + u(p^2 + \ldots + p^{d-4})}{p+1} \\
 & = u + p^2\cdot \frac{u(1+p+\ldots + p^{d-4})}{p+1} \\
 & = u + p^2\cdot \left( \frac{u(1+p) + u(p^2 + \ldots p^{d-4})}{p+1} \right) \\
 & = u + p^2\cdot \left( u + p^2\cdot \frac{u(1+p+\ldots p^{d-6})}{p+1} \right) \\
 & \quad \qquad\qquad \vdots \\
 & = u + p^2\cdot \left( u + p^2 \left( u + p^2\left( u + \ldots + p^2 \frac{u(1+p)}{1+p} \right) \right) \right) \\
 & = u + p^2\cdot ( u + p^2 \cdot ( u + p^2 ( u + \ldots + p^2 u ))) \\
 & \equiv u \pmod{p} \\
 & \equiv 1 \pmod{p}
\end{align*} 
and similar if $d-1$ is odd then successively multiply $(1+p+\ldots + p^{d-1})$ out with $v := (1+p+\ldots p^{d-2})$. But as said, this seems to complicated for me, so is there another easy way to see this?

There are even easier proofs than the answers others have supplied. If you look at the numerator, one of the terms is guaranteed to be divisible by $p^2-1$, whichever has an even exponent. This is because $(x^2-1)|(x^{2n}-1)$. The division clearly results in a polynomial with constant term $1$, as does the division of the other term by $p-1$. Then their product does as well, and so taking the expression mod p gives 1.
Alternatively, we observe that 
$p^d=0$ (mod p) and then just be done. This means that the fraction trivially simplifies to $$\frac{(-1)^2}{(-1)^2}=1$$
