Use the binomial theorem to prove that $\lim\limits_{n \to \infty} \sqrt[n]{c}$ converges to 1 for any c > 0. I will start following an Analysis 1 course from next week on, and I'm reading into the subject a bit. There's an exercise in the book that I can't solve. A similar one follows that exercise and they seem pretty much related.
This is really more about the formal epsilon definition of convergence than that of limits. The chapter isn't on limits, but on convergence (although they are closely related).
Use the binomial theorem to prove that $\lim\limits_{n \to \infty} \sqrt[n]{c}$ converges to 1 for any c > 0.
and 
Use the binomial theorem to prove that $\lim\limits_{n \to \infty} \sqrt[n]{n}$ converges to 1.
Note that this is the first ever chapter about series (and convergence) I've ever come across, and that it's therefore introductory. So there should be an 'easy' solution to the question.
EDIT: I've come as far as:
We want to prove there is an $n$* so that for any $n>=n$*, 
$|\sqrt[n]{c}-1|<\epsilon$ for any $\epsilon>0$.
Consider the case of c>1.
Then $|\sqrt[n]{c}-1|=\sqrt[n]{c}-1<\epsilon$
So
$\sqrt[n]{c}<\epsilon+1$
$c<(\epsilon+1)^n$
Now since $n\epsilon<(\epsilon+1)^n$ by the binomial theorem (I skip this proof for now), if 
$c<n\epsilon$ our statement still holds.
That is if
$n>c/\epsilon$
So consider the choice of $n$*$>c/\epsilon$.
Now this should be the so-called scratch-work. I don't know how to prove it, because inserting $n$* into the equation is gonna get really messy. 
Also I wonder if I made no assumptions I shouldn't make, did no things that are plain wrong, or if I took a route that's way too hard.
Thanks in advance.
 A: EDIT: I've noticed that you have changed your question to require "$\epsilon-\delta$" proof "from scratch". Just use the same inequalities proved below:
$1\le\sqrt[n]{c}\le 1+\frac{c}{n}$
$1\le\sqrt[n]{n}\le 1+\sqrt\frac{2}{n-1}$
Then for arbitrary $\epsilon>0$ find $n_0$ and $n_1$ such as $\frac{c}{n}<\epsilon$ for $n>n_0$ and $\sqrt\frac{2}{n-1}<\epsilon$ for $n>n_1$
Then you will get 
$|\sqrt[n]{c}-1|<\epsilon$ respectively $|\sqrt[n]{n}-1|<\epsilon$
OLD POST:
First part:
Write $\sqrt[n]{c}=1+b_n$, with $b_n>0$
Raise to the power $n$ and use Binomial Theorem:
$c=(1+b_n)^n=1+{n\choose 1}b_n+{n\choose 2}b_n^2+...+{n\choose n}b_n^n\ge {n\choose 1}b_n=nb_n$
since all the terms ${n\choose k}b_n^k$ on the right side are positive.
We get $n b_n\le c$ from where:
$0\le b_n\le\frac{c}{n}\rightarrow 0$
Second part:
Write $\sqrt[n]{n}=1+a_n$, with $a_n>0$
Raise to the power $n$ and use Binomial Theorem:
$n=(1+a_n)^n=1+{n\choose 1}a_n+{n\choose 2}a_n^2+...+{n\choose n}a_n^n\ge {n\choose 2}a_n^2$
since all the terms ${n\choose k}a_n^k$ on the right side are positive.
We get $\frac{n(n-1)}{2}a_n^2\le n$ from where:
$0\le a_n\le\sqrt\frac{2}{n-1}\rightarrow 0$
A: Hint:
$$n=\left(\sqrt[n]{n}\right)^n
=\left(1+\left(\sqrt[n]n-1\right)\right)^n
=\sum_{k=0}^n\binom{n}{k}\left(\sqrt[n]{n}-1\right)^k \geqslant\binom n2\left(\sqrt[n]{n}-1\right)^2 \\\,\\ $$
Then  We have  $$n \ge \binom n2 (\sqrt[n]{n} - 1)^2  \iff1 \le\sqrt[n]{n} \le 1+ \sqrt{\frac 2{n-1}}$$
Now its time to use Sandwich Theorem to prove $\lim\limits_{n \to \infty} \sqrt[n]{n}=1 $
A: For any $\epsilon\ge0$, the first two terms of the Binomial Theorem say
$$
1+n\epsilon\le(1+\epsilon)^n\tag{1}
$$
Multiply both sides of $(1)$ by $\frac{(1-\epsilon)^n}{1+n\epsilon}$ and we get
$$
\begin{align}
(1-\epsilon)^n
&\le\frac{\left(1-\epsilon^2\right)^n}{1+n\epsilon}\\
&\le\frac1{1+n\epsilon}\tag{2}
\end{align}
$$
For $n\ge\frac{c+\frac1c}\epsilon$, we have
$$
1-\epsilon\le\left(\frac1{1+n\epsilon}\right)^{1/n}\le c^{1/n}\le(1+n\epsilon)^{1/n}\le1+\epsilon\tag{3}
$$
Taking the limit, we get
$$
1-\epsilon\le\liminf_{n\to\infty}c^{1/n}\le\limsup_{n\to\infty}c^{1/n}\le1+\epsilon\tag{4}
$$
Since $(4)$ holds for any $\epsilon\gt0$, we must have
$$
\lim_{n\to\infty}c^{1/n}=1\tag{5}
$$

Extending the Result
Since $1+x\le e^x$, we have that $\left(1+\frac1n\right)^n\le e$. Therefore,
$$
\begin{align}
\frac{(n+1)^n}{n^{n+1}}
&=\frac1n\left(1+\frac1n\right)^n\\
&\le\frac en\tag{6}
\end{align}
$$
Thus, for $n\ge3$, we have that $n^{\large\frac1n}$ is a decreasing function, bounded below by $1$. This means that $\lim\limits_{n\to\infty}n^{\large\frac1n}$ exists.
Since we have $(5)$, we can proceed as follows
$$
\begin{align}
\lim_{n\to\infty}n^{\large\frac1n}
&=\lim_{n\to\infty}(2n)^{\large\frac1{2n}}\tag{7a}\\
&=\lim_{n\to\infty}2^{\large\frac1{2n}}\lim_{n\to\infty}n^{\large\frac1{2n}}\tag{7b}\\
&=\lim_{n\to\infty}2^{\large\frac1n}\tag{7c}\\[6pt]
&=1\tag{7d}
\end{align}
$$
Explanation:
$\text{(7a)}$: substitute $n\mapsto2n$
$\text{(7b)}$: product of limits is the limit of the products
$\text{(7c)}$: square $\text{(7b)}$ and divide by the left side of $\text{(7a)}$
$\text{(7d)}$: apply $(5)$
