I'm reading Vinberg's A course in Algebra and came across this question.
It is proved for a group $\mathbb Zp$ (p prime), that $(a+b)^p = a^p+b^p$.
The question is, Conclude from the aforesaid that every $a\in \mathbb Z_p$ satisfies the identity $a = a^p$.
Here is my proof by induction. I need to check whether my approach is correct. Specially when proving for the negative integers.
First, I prove the case for $n=1$.
From the above statement I have, $(a+b)^p = a^p+b^p$ for some $a\neq0$. Let $b=0$. Now, $(1a)^p = 1a^p$, This implies $1^p=1$ which is obvious.
Now, we assume $k^p=k$ for some $k\in \mathbb N$. Now set $b=ka$ which results in $(a+ka)^p=a^p+(ka)^p$. Since $k^p = k$ the above simplifies to $(k+1)^p\times a^p=a^p+ka^p.$ This implies, $(k+1)^p=1+k.$
By the principle of mathematical induction, we can show that $n = n^p$ for any natural number $n$.
But we need the proof for any integer (for negative integers too). So set $n=-a$ and repeat the above induction proof.
The case $a=0$ is trivial. Therefore, we can conclude that $a = a^p$ for any integer $a$.
Your proof seems unnecessarily complicated.
We have the base case
Assuming it's true for $k$. Then $k^p\cong k\pmod p$.
Then $(k+1)^p\cong k^p+1^p\cong k+1\square$.