Ok! I'll write $\displaystyle \mathbb{N}^*$ instead of $\displaystyle \mathbb{N}-\{0\}.$
We have $\displaystyle P(n):\ n\in\mathbb{N}^*\Rightarrow n^2\geq n$
Thus $\displaystyle P(1):\ 1\in \mathbb{N}^*\Rightarrow 1^2\geq 1$
Consider the formula $\displaystyle F:\ A\Rightarrow B,$ if $\displaystyle A$ is false or if $\displaystyle B$ is true, then $\displaystyle F$ is true.
$\displaystyle 1^2\geq 1$ is true, therefore the implication $\displaystyle 1\in \mathbb{N}^*\Rightarrow 1^2\geq 1$ is true: we've proved $\displaystyle P(1).$
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Another way to prove $\displaystyle P(1)$ (and even more) is to say that the proposition $\displaystyle P(x)$ is a tautology (i.e. is always true). Indeed:
$\displaystyle \circ$ If $\displaystyle x$ isn't a positive integer, then $\displaystyle x\in\mathbb{N}^*$ is false and the implication $\displaystyle x\in\mathbb{N}^*\Rightarrow x^2\geq x$ is true: $\displaystyle P(x)$ is true.
$\displaystyle \circ$ If $\displaystyle x$ is a positive integer, $\displaystyle (x\geq x\ \text{and}\ x\geq 1)\Rightarrow x\times x\geq x\times 1\Rightarrow x^2\geq x$ because the multiplication on $\displaystyle \mathbb{N}$ is compatible with the order. So the second part of the implication is true, and again $\displaystyle P(x)$ is true.
In conclusion, whatever is $\displaystyle x$ (a number, a matrix, a topological space...), $\displaystyle P(x)$ is true. In particular, $\displaystyle P(1)$ is true.