A non-exponential proof of $\lim_{x \to \infty} c^{1/x} = 1$ when $c>1$ My textbook asks me to prove this when we are just beginning to learn limits. 
$\lim \limits_{x \to \infty} c^{1/x} = 1, \  \ \ \ \ c>1$

Suppose otherwise,
$$\forall N > 0, \exists x, \epsilon \ | \ \ \ \ \ \ \ x> N \ \ \ \ \
 \Rightarrow \ \ \ \ \ \ \ \ \ c^{1/x} \geq 1 + \epsilon $$
$$ \begin{equation} \begin{split} &c \geq (1 + \epsilon)^x \ \ \ \ \ \
 \ \ \ &&\text{since $c>0$ } \\  &c \geq (1 + \epsilon)^N \ \ \ \ \ \ \
 \ \ &&\text{since $c^{x} > c^{N}$}  
\end{split} \end{equation}$$

Now this is obviously false, but I do not know how would you would prove it without a proof relying on a exponential function.
The best I could come up with is this:

Prove that for any $x,y > 1$ where $y > x$, there exists some $N$ such
  that $x^N > y$.
For any such $y$, there exists $z > y$. Let $x^N = y$. But again, this
  assumes that such an $N$ exists.

Could you suggest alternative ways of proving this? Or do you think that what I've done is satisfactory?

$$c^{1/x}$$ is a decreasing function of $x$, and it is bounded below (at least by $0$). So it converges.
Now if the limit is $L$,
$$L=\lim_{x\to\infty}c^{1/x}$$ implies $$L^2=\left(\lim_{x\to\infty}c^{1/x}\right)^2=\lim_{x\to\infty}c^{2/x}=\lim_{x\to\infty}c^{1/x}=L.$$ 
