Is this a good enough proof by induction? I am trying to understand induction. Is the proof good enough to satisfy what was looked for? If not, do you have any feedback for me?

To show: $\sum_{i=0}^n i^2 ≤ n^3$ for any $n = 0, 1, 2,\ldots$

If $P(n)$ holds then $P(n+1)$ holds as well $\sum_{i=0}^{n+1} i^2 ≤ (n+1)^3$
$$\sum_{i=0}^{n+1} i^2 = \sum_{i=0}^n i^2 + (n+1)^2 ≤ n^3 + (n+1)^2 ≤ (n+1)^3$$
$$\sum_{i=0}^n i^2 + (n+1)^2 ≤ n^3 + 3n^2 + 3n + 1$$
By the inductive hypothesis $\sum_{i=0}^n i^2 ≤ n^3$
$(n+1)^2 ≤ 3n^2 + 3n + 1$
$3n+1 ≤ 3n^2$ and
$(n+1)^2 ≤ 3n^2$
Thus, holds for $n+1$ and proof of induction is complete.
Thanks in advance.
 A: Although the proof seems right, it is not written correctly. Mainly it seems like you assume the claim throughout (even though you don't). Following would be a better way to write it:
Assume $P(n)$ is true, that is $\sum_{i=0}^n i^2 \leq n^3.$ We want to show $\sum_{i=0}^{n+1} i^2 \leq (n+1)^3.$
Consider
\begin{align*}
\sum_{i=0}^{n+1} i^2 &= \sum_{i=0}^{n} i^2 + (n+1)^2\\
&\leq n^3 + (n+1)^2\hspace{1cm} \text{(by induction hypothesis)}\\
&= n^3 + n^2+2n+1\\
&\leq n^3 +3n^2+3n+1\\
&=(n+1)^3. 
\end{align*}
Hence $P(n+1)$ holds.
A: Sahiba Arora’s answer shows how to write the induction step more clearly.  But besides that you also need to show the base case $P(0)$.  So I would write the whole proof something like:
Proof.  We work by induction on $n$.
The base case $P(0)$ states that $\sum_{i=0}^0 i^2 \leq 0^2$, i.e. $0 \leq 0$, which certainly holds.
For the induction step, assume $P(n)$ is true… [continued as in Sahiba Arora’s excellent answer].
So by induction, the given inequality holds for all natural numbers $n$.
