Lemma 1   For all integers $n$ greater than or equal to $5$ $$ 2^n > n^2 $$

Proof. We begin with a few easy observations. First, a bit of arithmetic: $$ 2^5 = 32 > 25 = 5^2 $$ Second, some algebra: $$ (n+1)^2 = 2 n^2 - (n-1)^2 + 2 $$ Third, a little more algebra: $$ n^2 - (n-1)^2 + 2 = 2n + 1 $$ This implies that $n^2 > (n-1)^2 - 2$ when $n > 0$ Also note that $(n-1)^2 > 2$ when $n \ge 2$

These observations provide us with the makings of an inductive proof. Suppose that $2^n > n^2$ for some integer $n > 2$ By our third observation and our hypothesis, $$ 2^n > n^2 - (n+1)^2 + 2. $$ Hence, $$ 2^{n+1} = 2^n + 2^n > 2 n^2 - (n+1)^2 + 2. $$ By the third observation above, the right-hand side equals $(n+1)^2$ The first observation above states that $2^n > n^2$ when $n = 5$ The reasoning presented above shows that $2^n > n^2$ implies $2^{n+1} > (n+1)^2$ when $n \ge 5$ Together, these assertions imply that $2^n > n^2$ when $n \ge 5$

Lemma 2   For all integers $n$ greater than or equal to $3$ $$ n^{n+1} > (n+1)^n $$

Proof. We begin by noting that $$ 3^4 = 81 > 64 = 4^3. $$ Next, we make assume that $$ (n-1)^n > n^{(n-1)}. $$ for some $n$ Multiplying both sides by $n$ $$ n (n-1)^n > n^n. $$ Multiplying both sides by $(n+1)^n$ and making use of the identity $(n+1)(n-1) = n^2 - 1$ $$ n (n^2 - 1)^n > n^n (n +1)^n. $$ Since $n^2 - 1 > n^2$ the left-hand side is less than $n^{2n+1}$ hence $$ n^{2n+1} > n^n (n +1)^n. $$ Canceling $n^n$ from both sides, $$ n^{(n+1)} > (n+1)^n. $$ Hence, by induction, $n^{(n+1)} > (n+1)^n$ for all $n \ge 3$

Theorem 1   $$ \lim_{n \to \infty} n^{1/n} = 1 $$

Proof. Consider the subsequence where $n$ is a power of $2$ We then have $$ (2^m)^{(1/2^m)} = 2^{m/2^m}. $$ By lemma 1, $m/2^m < 1/m$ when $m \ge 5$ Hence, $(2^m)^{1/2^m} < 2^{1/m}$ Since $\lim_{m \to 0} 2^{1/m} = 1$ and $(2^m)^{1/2^m)} > 1$ we conclude by the squeeze rule that $$ \lim_{m \to 0} (2^m)^{1/2^m} = 1. $$

By lemma 2, the sequence $\{n^{1/n}\}$ is decreasing. It is clearly bounded from below by $1$ Above, we exhibited a subsequence which tends towards $1$ Thus it follows that $$ \lim_{n \to \infty} n^{1/n} = 1. $$