quantum field theory - Assumptions in the LSZ reduction formula

In Srednicki, Chapter 5, it is said that the LSZ reduction formula holds only under the assumptions $$ \langle 0|\phi(x)|0\rangle =0 \qquad\text{and} \qquad \langle k|\phi(x)|0\rangle =e^{-ikx}$$

I don't understand why the first assumption (and a related one discussed in the chapter) is necessary. It seems that the assumption follows from Lorentz invariance.

We know, using only the fact that $\phi(x)$ is a Lorentz scalar, that $$ [P^{\mu},\phi(x)]=i\partial ^{\mu}\phi(x)$$

Then, we have $$ a^{\dagger}(k)=-i\int d^3x \; e^{ikx}[-i\partial_0\phi(x)+\omega \phi(x)]$$

Therefore, \begin{equation} [P^{\mu},a^{\dagger}(k)] = -i\int d^3x \; e^{ikx}[\partial_0\partial^{\mu}\phi(x)+i\omega \partial^{\mu}\phi(x)]\\ =\int d^3x \; e^{ikx}\partial^{\mu}[-i\partial_0\phi(x)+\omega \phi(x)]\\ =-\int d^3x \; \partial^{\mu}(e^{ikx})[-i\partial_0\phi(x)+\omega \phi(x)]\\ =-ik^{\mu}\int d^3x \; e^{ikx}[-i\partial_0\phi(x)+\omega \phi(x)]\\ =k^{\mu} a^{\dagger}(k) \end{equation}

This shows that $a^{\dagger}(k)|0\rangle$ is now an eigenvector of $P^{\mu}$ of eigenvalue $k^{\mu}$. So we have: $$ a^{\dagger}(k)|0\rangle = \lambda |k\rangle $$

It seems to me that it has two implications:

One is that $$ \langle 0|a^{\dagger}(k)|0\rangle = 0$$ and hence $$ \langle 0|a(k)|0\rangle = \overline{\langle 0|a^{\dagger}(k)|0\rangle}=0$$

Then, doesn't it imply $$\langle 0 | \phi(x) | 0 \rangle = 0$$ only from the assumption that $\phi(x)$ is a Lorentz scalar? (Without any renormalization assumption)

The second consequence is that $$ \langle k',n|a^{\dagger}(k)|0\rangle = 0$$ for any $k'\neq k$.

If so, why did Srednicki spend so much effort to prove that $$ \langle k',n|a_1^{\dagger}(t)|0\rangle \rightarrow 0\qquad \text{as}\qquad t\rightarrow \pm \infty$$