Show that $\displaystyle 1<\sqrt[n]{a}<\frac {n+(a-1)}n$ if $\displaystyle a>1$ and $\displaystyle n\in\mathbb{N}\setminus\{1\}$.
That $\displaystyle \sqrt[n]{a}>1$ is easy to see. For the second inequality, $\displaystyle \sqrt[n]{a}<\frac {n+(a-1)}n\iff a<(1+(a-1)/n)^n$. The latter can be shown using the binomial theorem, according to which $\displaystyle (1+x)^n=1+nx+\dots>1+x$.
In fact, the whole power of the theorem is not needed. One can write $\displaystyle (1+x)^n$ as $\displaystyle (1+x)\cdot(1+x)\cdot\;\dots\;\cdot(1+x)$. When performing the multiplication, we have to pick either 1 or x from each factor, and consider and add all such variants. There is only one way to choose 1 from every factor; all other terms will have x in them. Then we can choose x from the first factor and 1's from all the rest, or x from the second factor and 1's from all the rest, etc. In the sum, this will give us $\displaystyle nx$. All the rest terms in the sum will have at leat $\displaystyle x^2$.