$phi^4$-theory, S-matrix Feynman diagram to first order from Peskin and Schroeder

Suppose we have a general term of the form $$langle 0 | a_{p_1} a_{p_2} cdots a_{p_{n-1}}a_{p_n} a_{p_1}^dagger a_{p_2}^dagger cdots a^dagger_{p_{m-1}}a_{p_m}^dagger | 0 rangle$$ We can eliminate $a_{p_1}^dagger$ by commuting it past all the $a_{p_i}$ factors until it anhilitates the vacuum state. Along the way, we will end up with a bunch of terms of the form $$langle 0 | a_{p_1} cdots [a_{p_i},a_{p_1}^dagger] cdots a_{p_n} a_{p_2}^dagger cdots a_{p_m}^dagger | 0 rangle.$$ In each of these terms, we can use the commutation relation $$[a_p,a_{p'}^dagger]=(2pi)^3delta^{(3)}(p-p')$$ to eliminate $a_{p_i}$ and $a_{p_1}^dagger$ from the term. The net result is the sum of a bunch of terms, each with one less $a$ and one less $a^dagger$. We can continue this process until we run out of either $a^dagger$ factors or $a$ factors.

At this point, there are three possibilities. If there were more $a$ factors than $a^dagger$ factors, we will have the sum of a bunch of terms, each the product of delta functions with a factor of the form $langle 0|a_{p'_1}cdots a_{p'_{n-m}}|0rangle$. These terms are all zero because an annihilation operator appears next to $|0rangle$. Similarly, if there were more factors of $a^dagger$, we would end up with an $a^dagger$ next to $langle 0|$ in each term. The only way to avoid a trivial result is if there are an equal number of factors of $a$ and $a^dagger$. Then we will end up with a sum of products of delta functions.