Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$

Your argument is fine and quite clearly presented. You can shorten the presentation considerably, though, by doing something like this:

For $n\ge 2$ let $a_n=\sum_{k=n}^{n^2}\frac1k$. Since $a_2=\frac12+\frac13+\frac14>1$, it suffices to show that $a_{n+1}\ge a_n$ for $n\ge 2$. Since $n^2+2n+1<2n^2+n$ for $n\ge 2$, we have

$$a_{n+1}-a_n=\sum_{k=1}^{2n+1}\frac1{n^2+k}-\frac1n\ge\sum_{k=1}^{2n+1}\frac1{2n^2+n}-\frac1n=\frac{2n+1}{2n^2+n}-\frac1n=0\;,$$

so $a_{n+1}\ge a_n$, and the result follows by induction.

If you accept a proof without induction, here is one: \begin{align*} \frac 1n+\Bigl(\frac 1{n+1}+\dots+ \frac1{n^2}\Bigr)>\frac 1n +(n^2-n)\cdot\frac 1{n^2}=\frac 1n+1-\frac1n=1. \end{align*} This computation is valid if $n^2>n$, i.e. if n>1.