Prove using induction that ${p - 1 \choose n} \equiv (-1)^n \bmod p$ for $0 \leq n \leq p - 1$ where $p > 2$ is prime Can someone see if this method is correct because I have doubts using the theory of induction?

Let $p > 2$ be a prime number. Prove that
\begin{align*} \binom{p-1}{n} \equiv (-1)^n \, (\mathrm{mod} \, {p}) \;  \text{for any number }  \ 0 \leq n \leq p-1  \end{align*}

My Attempt
\begin{align*}
\binom{p}{n} = \binom{p-1}{n} + \binom{p-1}{n-1} 
\end{align*}
Is divisible by $p$. Therefore,
\begin{align*}
\binom{p-1}{n} = - \binom{p-1}{n-1} \pmod p
\end{align*}
Hence, for the induction step,
\begin{align*}
\binom{p-1}{n-1} \equiv (-1)^{n-1} \pmod p
\end{align*}
It follows that
\begin{align*}
\binom{p-1}{n} \equiv (-1)^n \pmod p
\end{align*}
It was the first time I wrote in MathJax.
It took me a long time... but here goes with practice. :)
Thanks in advance.
 A: Your approach is correct, as you fix $p$ to be a fixed prime number, and attempt to prove for all $0\leq n\leq p-1$ that;
$$\binom{p}{n}=(-1)^n\pmod{p}.$$
In this approach, however, we need to first prove that the claim does hold for $n=0$ and $n=1$ which are the Base Cases. Once we have proved the statement for $n=0$ and $1$, we can approach the Inductive Step; to prove that if the statement is true for $n-1$ and $n$ in $\{0,1,2,\dots, p-2\},$ then it is also true for $n+1.$
Once we have done this, we can see the dominoes falling as the claim gets proven for all of the required numbers;
$$\text{Claim true for }n=0\text{ and }1\implies \text{So claim also true for }n=2$$
$$\text{Claim true for }n=1\text{ and }2\implies \text{So claim also true for }n=3$$
$$\text{Claim true for }n=2\text{ and }3\implies \text{So claim also true for }n=4$$
$$\vdots$$
$$\text{Claim true for }n=p-3\text{ and }p-2\implies \text{So claim also true for }n=p-1.$$
Finally, we Can see its true for all of the required values of $n.$ Hope this method works out for you.
