This problem is from An Introduction to Proof through Real Analysis by D. J. Madden and J. A. Aubrey.
Prove that for all natural numbers $n$, $n^2 \geq n$.
I propose two different proofs.
Proof 1. This proof is induction on $n$.
We let $n = 1$, then $n^2 = 1 \cdot 1 = 1 = n$. So the statement is true for $n = 1$.
Now we let $n = n_0$, and assume that the statement holds. So $n_0^2 \geq n_0$.
Let $n = n_0 + 1$, then we have \begin{align} (n_0 + 1)^2 = n_0^2 + 2n_0 + 1 &\geq n_0 + 1 \\ n_0^2 + 2n_0 &\geq n_0 \tag{cancellation of addition} \\ n_0^2 + n_0 &\geq 0 \tag{cancellation of addition}. \end{align} Which is true. So the statement holds for $n_0 + 1$. Hence the statement is true for all natural numbers.
Proof 2. We use the same information and verify that it works for $n = 1$. Now for $n > 1$, we can express $n$ as an eventual successor of $1$, which is $$n = ((( \cdots (1 + 1)+1)+1)$$ By the definition of multiplication,
$n = ((( \cdots (1 + 1) + 1) + 1) + 1).$
$n \cdot m = ((( \cdots (m + m) + m) + m) + m).$
So $n^2 \equiv n \cdot n$ can be written as \begin{align} n \cdot n &= ((( \cdots (n + n) + n) + n) + n) \\ &= ((( \cdots (n + ((( \cdots (1 + 1) + 1) + 1) + 1)) + n) + n) + n) \\ &= \vdots \end{align} which is obviously larger than $n$, because $n^2$ can be written as a successor of $n$. Hence, we claim that for $n > 1$, $n^2 > n$. The logical union of these two cases is $n^2 \geq n$.
Are my proofs valid? If they aren't, which parts do not make sense? Thank you.
Your ideas appear to be correct, but the way you have written up the proofs make them rather non-understandable, and you need to fix your presentation and logic.
In proof 1, the idea of induction is: to prove $P(n)$ for all positive integers $n$, we show $P(1)$, and then show $P(n)\implies P(n+1)$ for all $n$. You have shown $P(1)$ correctly, very nice. But your logic in showing that $P(n)\implies P(n+1)$ is faulty. In particular, $(n_0+1)^2\geq n_0+1$ is what you need to show, but you have started with that statement without explanation, so your proof is not correct. If you instead note that each line is an equivalent statement, that is, $$\begin{split}&(n_0+1)^2\geq n_0+1\\\iff &n_0^2+2n_0\geq n_0\\\iff& n_0^2+n_0\geq0 \end{split}$$ and then observe the last line is true, then you will in fact have a correct proof, becuse then the chain of logical implications goes back up: starting from a true statement you deduce back the true first statement.
But realise that in your entire proof, you did not ever use the inductive hypothesis that $n_0^2\geq n_0$. So it is not even an induction proof at all. In reality your proof $1$ should be written more succintly as follows: $$n^2\geq n\iff n^2-n\geq 0\iff n(n-1)\geq0$$ which is true since $n\geq 1$, so both $n$ and $n-1$ are $\geq 0$.
In proof 2, you might have the right idea, but writing an integer as the sum of a bunch of $1$s only obfuscates whatever you are trying to say. There is literally zero point in doing this. What you are really trying to say is that for $n\geq 2$, $$n^2=n\cdot n\geq 2n>n.$$
In conclusion, both proofs might have roughly the right idea, but the way you wrote it makes the ideas difficult if not impossible to discern. So I would say your proofs are both incorrect.