Show that the orthogonal projection onto $Range T$ is equal to $T(T^\ast T)^{-1}T^\ast$ given that $T: V \to W$ is injective

First, note that if $T^{*}Tv = 0$ then

$$ 0 = \left< T^{*}Tv, v \right> = \left < Tv, Tv \right> = \| Tv \|^2 = 0 $$

so $Tv = 0$. Since $T$ is injective, it implies that $v = 0$. Hence, $T^{*}T \colon V \rightarrow V$ is invertible. Now,

$$ P^2 = T(T^{*}T)^{-1}T^{*} T(T^{*}T)^{-1} T^{*} = T(T^{*}T)^{-1}T^{*} = P $$

and

$$ P^{*} = \left( T(T^{*}T)^{-1}T^{*} \right)^{*} = \left( T^{*} \right)^{*} \left( (T^{*}T)^{-1} \right)^{*} T^{*} = \\T \left( (T^{*}T)^{*} \right)^{-1} T^{*} = T \left( T^{*} \left (T^{*} \right)^{*} \right)^{-1} T^{*} = T(T^{*}T)^{-1}T^{*} = P$$

so $P$ is self-adjoint. Hence, $P$ is an orthogonal projection onto $\operatorname{Range}(P)$. Finally, since $T$ is injective, we have that $T^{*}$ is surjective and since $(T^{*}T)$ is an isomorphism we have

$$ \operatorname{Range}(P) = \operatorname{Range}(T). $$

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Let's see how you can arrive the formula intuitively. Since $P$ is an orthogonal projection on $\operatorname{Range}(T)$, it should vanish on $\operatorname{Range}(T)^{\perp} = \ker(T^{*})$ so it makes sense to look for $P$ of the form $P = AT^{*}$. You also want that the range of $P$ will be the range of $T$ so that it makes sense to look for $P$ of the form $P = TAT^{*}$ with $A \colon V \rightarrow V$ being an isomorphism. In order for $P$ to be a projection, we should have

$$ P^2 = T(AT^{*}T)AT^{*} = TAT^{*} = P$$

which will hold if $AT^{*}T = \operatorname{id}|_{V}$ or $A = (T^{*}T)^{-1}$.