Is $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ correctly proved by induction? I have this statement:

Prove by induction that $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ for $n \in \mathbb{N}$

My attempt was:
Base case: $\frac{1}{1} \leq 1$
Assume that $\displaystyle \sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$ is true for some $m \in \mathbb{N}$
To prove: $\displaystyle \sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$
Since $$\sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$$
$$\sum_{k = 1}^m\frac{1}{k^2}  + \frac{1}{(m+1)^2}\leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$
$$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$
Now, I'll prove that (1) $$\displaystyle 2 -\frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m + 1}$$
Developing this inequation, gives $$m^2 + 2m + 1 \geq m^2 + 2m$$ getting that (1) is true.
And by transitivity $$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$$ was proved.
But I don't know if this is a valid induction proof and if not, what is the type of this proof?
Thanks in advance.

If we know that it is valid for $n$, we only need to show that $\frac{1}{(n+1)^{2}}<\frac{1}{n}-\frac{1}{n+1}$
Therefore
$$
\sum_{k=1}^{n}{\frac{1}{k^{2}}}+\frac{1}{(n+1)^{2}}<2-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}
$$
