$e^z=z$ has infinitely many solutions

Let \[ F(z)=e^z-z. \] This is an entire function of order at most \(1\). Suppose, for contradiction, that it had only finitely many zeros. Hadamard’s factorization theorem would then give \[ F(z)=e^{az+b}Q(z) \] for some constants \(a,b\in\mathbb C\) and a polynomial \(Q\) whose zeros are exactly those finitely many zeros of \(F\).

Along the negative real axis, \[ F(x)=e^x-x\sim -x\qquad(x\to-\infty), \] so \[ \log|F(x)|=\log|x|+o(1). \tag{1} \] If \(m=\deg Q\), the factorization gives \[ \log|F(x)|=(\Re a)x+m\log|x|+O(1). \tag{2} \] Comparing (1) and (2) forces \[ \Re a=0, \qquad m=1. \tag{3} \]

On the positive real axis, however, \[ F(x)=e^x-x\sim e^x, \] so \[ \log|F(x)|=x+o(1). \tag{4} \] But (2) and (3) give only \[ \log|F(x)|=\log x+O(1), \] contradicting (4). Therefore \(e^z-z\) has infinitely many zeros in \(\mathbb C\).