Here, $a_n:=n^{\frac{1}{n}}-1$.
My attempt has been to write
\begin{align*}(1+a_n)^n =& \sum_i^n\binom{n}{i}a_n^i\\ &\geq 1+na_n + \frac{n(n-1)a_n^2}{2}. \end{align*} I can't see how to proceed from here. I could somehow show that the LHS is less than or equal to $n+na_n+1$, I'd be done, as the result would be implied.
The question has been solved in the comments, hence I will prove a sharper upper bound for $a_n=n^{1/n}-1$. For any $n\geq 2$ we have $n=\prod_{k=1}^{n-1}\frac{k+1}{k}$, hence:
$$ n^{1/n}=\text{GM}\left(1,\tfrac{2}{1},\ldots,\tfrac{n}{n-1}\right)\leq\text{AM}\left(1,\tfrac{2}{1},\ldots,\tfrac{n}{n-1}\right)=1+\frac{1}{n}\sum_{k=1}^{n-1}\frac{1}{k} \tag{1}$$ and by the Cauchy-Schwarz inequality $$ H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}\leq\sqrt{(n-1)\sum_{k=1}^{n-1}\frac{1}{k^2}}\leq\sqrt{\frac{\pi^2}{6}(n-1)}\tag{2} $$ hence by $(1)$ and $(2)$ we have that $$ \forall n\geq 2,\qquad a_n=n^{1/n}-1\leq\color{blue}{\sqrt{\frac{\zeta(2)}{n}}}.\tag{3}$$ By studying the function $g(x)=\sqrt{x}\left[\exp\left(\tfrac{\log x}{x}\right)-1\right]$ over $[1,+\infty)$ we have that $(3)$ can be improved up to $$ \forall n\geq 1,\qquad a_n=n^{1/n}-1 \leq \color{blue}{\sqrt{\frac{29}{34\,n}}}.\tag{4} $$
