Degree of a projected curve I cannot find the proof of the following fact, can anyone help me? 

Let $C$ be a projective curve in $\mathbb P^n$ and $p\in C$ a smooth point. Let $C'$ be the closure in $\mathbb P^{n-1}$ of the image of $C\setminus p$ via the projection from $p$. Prove that $\deg C' = \deg C-1$.

Thanks!
 A: I think one could argue as follows: Let $W$ be the variety of hyperplanes through $p \in C$. Then $\dim W = n-1$. Now for an arbitrary curve $D \subset \mathbf{P}^n$ consider the subset $\Gamma \subseteq D \times W$ defined by
$$\Gamma = \{ (x,H) \in D \times W \mid H \text{ tangential to } D \text{ at } x\}$$
Then the fiber over $x \in D$ of $\Gamma \to D$ has dimension $n-3$, therefore $\dim \Gamma = n-2$. As $\dim W = n-1$ the image of $\Gamma$ in $W$ has an open complement. Now let $D$ be first $C'$ giving $\Gamma_0=\Gamma_{C'}$.
Next let $D$ be $C - \{p\}$ giving $\Gamma_1$. Take $\Gamma_{11} = \bar{\Gamma_1}$, the closure of $\Gamma_1$ in $C \times W$. Additionally let $\Gamma_2$ be the subset of $W$ consisting of hyperplanes $H$ tangential to $C$ at $p$. Then $\dim \Gamma_0 = \dim \Gamma_{11} = \dim \Gamma_2 = n-2$.
So the complement of the image of the union of these three varieties in $W$ is an open set in $W$.
It consists of hyperplanes $H$ which are transversal to $C$ and $C'$ everywhere and so furnish a proof of the assertion. Of course $n$ must be greater than $2$.
