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?

I shall assume that we are taking the limit only over natrual $x$, i.e., we can write it more suggestively as
$$ \lim_{n\to\infty}c^{1/n}.$$
I shall also assume that you define $c^{1/n}$ as the unique positive real number $y$ such that $c=y^n$ (where the latter is just the product of $n$ identical factors).
As $0<y\le1$ implies $y^n\le1$, we conclude that $c^{1/n}>1$ for our $c>1$
Let $\epsilon>0$. We want to find $N$ such that $|c^{1/n}-1|<\epsilon$ for all $n>N$. 
By the archimedean property of the real numbers, there exists $N\in\Bbb N$ with $N>\frac c\epsilon$. 
Then we find that for all $n>N$, by the Bernoulli inequality,
$$ (1+\epsilon)^n\ge 1+n\epsilon>1+N\epsilon>1+c>c$$
and hence $1+\epsilon>c^{1/n}>1$ and ultimately $|c^{1/n}-1|<\epsilon$. (Note that we used $0<a<b\implies a^n<b^n$ somewhere along the way)
