Prove that $N = \frac{b^n-1}{b-1}$ is a pseudoprime number to the base b. Let $n$ a pseudoprime-number to the base $b$ and $\gcd(b - 1, n) = 1$
Prove that $N = \frac{b^n-1}{b-1}$ is a pseudoprime number to the base b.
My Atachment:
Proof. Note that $Φ_p(b) = M_p (b)$ and $\gcd(p, Φ_p(b)) = 1$.
So $P_p(b) = M_p(b)$
Let $N > 2$ and $P_N (b) = \frac{Φ_N (b)}{ gcd (N, ΦN (b))}$. 
If $P_N (b)$ is
composite, then $P_N (b)$ is an overpseudoprime to base $b$.
Source: https://arxiv.org/pdf/1206.0606.pdf

is that right?

I have an alternative approach.
We want to see that:
\begin{equation*}
    b^{N-1}\equiv 1 ( \text{ mod } N)
\end{equation*}
lets begin by proving that $n|N-1$.
$n$ is pseudoprime, i.e
$$b^{n-1} \equiv 1 ( \text{ mod } n)$$
Therefore $n|( b^{n-1} -1 )$ which is equivalent to say that $n| b^n-b$.
On the other hand, we know that
$N-1=\frac{b^n-1}{b-1}-1=\frac{b^n-b}{b-1}$
Now we use the fact that $gcd(b-1,n)=1$ to confirm that $(b-1)$ and $n$ don't have any common divisor, and since $n$ divides the numerator, he obtain  $n|N-1$.
We can express $N-1=n*k$.
Now we have:
\begin{equation*}
    b^{N-1}\equiv (b^{n})^k ( \text{ mod } N)
\end{equation*}
but $b^n \equiv 1 ($ mod $N)$, since
\begin{equation*}
    b^n-1\equiv N \cdot (b-1) \equiv 0 \cdot (b-1) \equiv 0 (\mod N)
\end{equation*}
Thus,
\begin{equation*}
    b^{N-1}\equiv (b^{n})^k\equiv 1^k\equiv 1 ( \text{ mod } N)
\end{equation*}
