Problem handling free groups in algebraic topology

Trying to compute the fundamental group of a topological space $X$ I have come to the equality $$\pi_{1}(X)\cong\frac{\ast_{i=1}^{n}\mathbb{Z}}{G}$$ where $\ast$ means taking the free product ($n$ times) of the group $\mathbb{Z}$ and $G$ is the smallest normal subgroup of $\ast_{i=1}^{n}\mathbb{Z}$ which contains the free group generated by the word $z_{1}z_{2}...z_{n}$ where each $z_{i}$ is the generator of a different $\mathbb{Z}$ of the product.

Now here is my problem:

It seems to me that this group is isomorphic to $\mathbb{Z}$, and I don't know wether $\mathbb{Z}$ is normal in the product, so I don't know if $\mathbb{Z}\cong G$, and even if that is the case I have no clue as to what is the quotient.

Could you please help me out?

Here

$$\begin{align} \langle z_1,z_2,\dots, z_n\mid z_1z_2\dots z_n\rangle &\cong \langle z_1,z_2,\dots, z_n\mid z_1=(z_2\dots z_n)^{-1}\rangle\\ &\cong\langle z_2,\dots, z_n\mid \rangle\tag{1}\\ &\cong\ast_{i=1}^{n-1}\Bbb Z, \end{align}$$

where $(1)$ holds by a simple Tietze transformation.