show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. I am trying to show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. The reverse implication is easy, but the forward implication is tricky. It seems essentially to state the the (n-1)th root of a natural number n is not an integer for n>2.
 A: Hint:
By Bernoulli inequality 
$$b^{c-1} \geq 1+(c-1)(b-1)$$
Now note that 
$$1+(c-1)(b-1) \leq c \Leftrightarrow \\
2+bc-b-2c \leq 0 \Leftrightarrow \\
(c-1)(b-2)  \leq 0$$
This leaves very few cases to check.
A: Define $f(x) = \dfrac{\ln x}{x-1}$. Then, $f'(x) = -\dfrac{1}{(x-1)^2}\ln x + \dfrac{1}{x-1} \cdot \dfrac{1}{x} = -\dfrac{x(\ln x - 1)+1}{x(x-1)^2}$. 
For $x \ge 3$, we have $f'(x) < 0$, so $f$ is decreasing on $[3,\infty)$. 
Hence, for all $c \ge 3$, we have $f(c) \le f(3)$, i.e. $\dfrac{\ln c}{c-1} \le \dfrac{\ln 3}{2}$. 
Exponentiating both sides to get $c^{\tfrac{1}{c-1}} \le \sqrt{3} < 2$. 
Trivially, $c^{\tfrac{1}{c-1}} > 1$ for integers $c \ge 3$. Therefore, $1 < c^{\tfrac{1}{c-1}} < 2$ for all integers $c \ge 3$. 
Hence, $c^{\tfrac{1}{c-1}}$ is not an integer for any integer $c \ge 3$.
A: As Joey notes we have  $b^{c−1}=c$. Let's rewrite this as $b^{d}=d+1$. The left-hand side grows much faster than the right. For $d>1, b>1$, we have $b^d=(1+n)^d = (1 + dn + ...)$ in the binomial expansion where the missing ... terms are $> 0$. But $1+dn+...> 1+d$. Equality is only possible if d=1, meaning c=2, and so b=2.
A: $1<  c = b^{c-1}\!\Rightarrow\,b\neq 1\Rightarrow\,b\ge 2\,\Rightarrow\,b^{c-1}\ge 2^{c-1} > c\,$ if $\,\color{#c00}{c > 2}\ $ (proof: $\,2^2 > 3\,$ and the inductive step is straightforward $\, 2^{c-1} > c\,\underset{\large \times 2\,}\Rightarrow\, 2^c> \color{#c00}{2c > c\!+\!2}> c\!+\!1)$
