Prove sequence $a_n=n^{1/n}$ is convergent

Noticing that $n^\frac{1}{n} > 1$ for all $n$, it all comes down to showing that for any $\epsilon > 0$, there is a $n$ such that $(1+\epsilon) \geq n^\frac{1}{n}$, or by rearranging, that

$$ (1+\epsilon)^n \geq n $$

Now, let's first of all choose an $m$ such that $(1+\epsilon)^{m}$ is some number bigger than 2, let's say the smallest number greater than $3$ that you can get. From here, swap $m$ for $2m$. This will make the left side a little over 3 times larger, and the right side 2 times larger. The next doubling will still double the right side, but the left side will increase roughly 9-fold. Repeating, we can easily see that the left side will at some point overtake the right side, and we have our $n$

To show $\lim\limits_{n \to \infty} n^{1/n} = 1$, we need to show that for all $\varepsilon > 0$, there exists $N \in \mathbb{N}$ s.t. $\mid n^\frac{1}{n} - 1 \mid < \varepsilon$ for any $n > N$.

Expand \begin{equation} \mid n^\frac{1}{n} - 1 \mid < \varepsilon \Leftrightarrow (1- \varepsilon)^n < n < (1+\varepsilon)^n \end{equation}

The left part holds as $(1- \varepsilon)^n < 1 \le n$ For the right part, we use the binomial theorem, $$ (1+\varepsilon)^n = \sum_{i=0}^n \binom{n}{i}\varepsilon^i > \binom{n}{2}\varepsilon^{2} = \frac{n(n-1)}{2} \varepsilon^2 $$ Set $N = \lceil \frac{2}{\varepsilon^2} + 1 \rceil$, we have $$ (1+\varepsilon)^n > \frac{n(n-1)}{2} \varepsilon^2 > n $$

So here is an outline of a proof:

Step 1: Notice that $n^\frac{1}{n}\geq 1$ for all $n$.

Step 2: Prove that $a_n$ is monotonically decreasing for $n\geq 3$. Equivalently we need to show that $n^{(n+1)}>(n+1)^n$.

Step 3: Show that there is a subsequence which converges to $1$. I managed to do this by considering $b_n={a_{2^{2^n}}}$. (It does not appear well in LaTeX as there are too many nested exponents. I had typed this part out, but decided to remove it)

From these three facts you can conclude that the limit is $1$.