Show that if $(b^n-1)/(b-1)$ is a power of prime numbers, where $b,n>1$ are positive integers, then $n$ must be a prime number. 
Show that if $(b^n-1)/(b-1)$ is the power of a prime number, where $b,n>1$ are positive integers, then $n$ must be a prime number.

My solution:
If $n$ is composite, then let $n=mk$, $m,k>1$,
\begin{align*}
\frac{b^n-1}{b-1} &= 1+b+\cdots+b^{n-1} \\
&=(1+b+\cdots+b^{k-1} )+(b^k+b^{k+1}+\cdots+b^{2k-1}) \\ &\quad\,+\cdots+(b^{(m-1)k}+b^{(m-1)k+1}+\cdots+b^{mk-1}) \\
&=(1+b+\cdots+b^{k-1})(1+b^k+\cdots+b^{(m-1)k})
\end{align*}
Which is composite and distinct, thus, for $(b^n-1)/(b-1)$ to be a power of primes, $n$ is not composite, thus it must be prime.
However, $(1+b+\cdots+b^{k-1})(1+b^k+\cdots+b^{(m-1)k})$ might be equal to $p^x \times p^y$, where $p$ is prime.
Is there any better solution?

Let $(b^n-1)/(b-1)=p^x$ where $p$ is a prime and $x> 0$. If $n$ is composite, there are two cases.


*

*There exists a prime $q$ such that $n=q^m$ for some $m>1$. Note 
$$p^x=\frac{b^n-1}{b-1}=\frac{b^{q^m}-1}{b^{q^{m-1}}-1}\cdot \frac{b^{q^{m-1}}-1}{b-1},$$
we can assume 
$$\frac{b^{q^{m-1}}-1}{b-1}=p^y$$
for some $0< y< x$. Then we have
\begin{align}
1+q(b-1)p^y+\sum_{i=2}^q\binom{q}{i}\left((b-1)p^y\right)^i&=\left((b-1)p^y+1\right)^q\\
&=\left(b^{q^{m-1}}\right)^q\\
&=b^{q^m}\\
&=(b-1)p^x+1,
\end{align}
i.e., 
$$q+\sum_{i=2}^q\binom{q}{i}\left((b-1)p^y\right)^{i-1}=p^{x-y}.$$
Hence, $p\mid q$. Recall that $p$ and $q$ are both primes, so $p=q$, we further have
$$1+\binom{p}{2}(b-1)p^{y-1}+\sum_{i=3}^p\binom{p}{i}(b-1)^{i-1}p^{y(i-2)}=p^{x-y-1}.$$
Note the left hand side is no less than 2, so both sides are divisible by $p$, i.e., the term $\binom{p}{2}(b-1)p^{y-1}$ cannot be divisible by $p$, thus $p=2$ and $y=1$. We further have $b=p^{x-2}$, i.e., $p\mid b$ (recall that $b>1$). However, note 
$$p^x=\frac{b^n-1}{b-1}=1+b+\cdots+b^{n-1},$$ 
it is impossible that $p\mid b$.

*There exist two co-prime numbers $s,t>1$ such that $n=st$. In this case, we have 
$$p^x=\frac{b^n-1}{b-1}=\frac{b^{st}-1}{b^s-1}\cdot\frac{b^s-1}{b-1},$$
which means $(b^s-1)/(b-1)$ is divisible by $p$. Similarly, $(b^t-1)/(b-1)$ is also divisible by $p$. Since $s$ and $t$ are co-prime, there exists integers $w_s,w_t$ such that $w_ss+w_tt=1$. Without loss of generality, we assume $w_s>0$ and $w_t<0$. Then we have
$$\frac{b^{w_ss}-1}{b^s-1}\cdot\frac{b^s-1}{b-1}-b\cdot\frac{b^{-w_tt}-1}{b^t-1}\cdot\frac{b^t-1}{b-1}$$
is also divisible by $p$. Note the expression above is exactly
$$\left(1+b+\cdots+b^{w_ss}\right)-b\left(1+b+\cdots+b^{-w_tt}\right)=1,$$ 
which is impossible.
As a conclusion, $n$ must be a prime.
