Prove that T-cI is normal

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Can anyone help me proving this: Let $(V,<,>)$ be a finite dimensional vector space , Let $T$ be a normal linear operator on $V$. Then $T-cI$ is normal for every $c \in \mathbb{F}$, $c \neq 0$

I have tried this, but I do not know how to get to the conclusion that T is normal

$(T-cI)(T-cI)^\ast=(T-cI)(T^\ast-\bar{c}I)=TT^\ast-\bar{c}IT-cIT^\ast+c\bar{c}I$

That is what I have tried

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I think your problem also probably wanted you to assume $T$ is normal to begin with. (Otherwise you are claiming literally every transformation is normal. If we assume the conclusion is true then for $c=0$ you would have had to started out with a normal operator.)

That being the case, when you write out both $(T-cI)(T-cI)^\ast$ and $(T-cI)^\ast(T-cI)$, the equation $TT^\ast=T^\ast T$ is going to help you switch the order of $T$ and its adjoint, so that you can get equality.

The real upshot here is that any polynomial in $T$ is going to commute with any other polynomial in $T^\ast$, and $X-cI$ is just one of those...