Given entire function has infinitely many zeros Prove that for any $\lambda \neq 0$ and any polynomial $p(z) \not\equiv 0,$ the function $g(z)=e^{\lambda z}-p(z)$ has infinitely many zeros.
My approach: Suppose to the contrary that $g$ has finitely many zeros, at $a_j$'s (say)
$$g(z)=\prod_{j=1}^{n} (z-a_j).$$
By Hadamard's factorization
$$g(z)=e^{h(z)} \cdot \prod_{j=1}^{n} (z-a_j),$$
for some polynomial $h(z).$ I don't know how to get a contradiction. Any help is much appreicated.
 A: The same approach as in Showing that $e^z=z$ has infinitely many solutions can be used:
$h(z) = p(z) e^{-\lambda z}$ has an essential singularity at $z = \infty$,
and has only finitely many zeros. Using Great Picard's Theorem it follows that 
$h$ takes any non-zero value infinitely often. In particular, 
$h(z) = 1$ has infinitely many solutions. 
This is the desired conclusion because $h(z) = 1 \Longleftrightarrow e^{\lambda z} = p(z)$.
A: An alternative proof: If the claim is false then
$$ \tag{*}
  e^{\lambda z} - p(z) = q(z) e^{h(z)}
$$
with non-zero polynomials $p, q$, and an entire function $h$. 
To simplify the notation let us assume that $\lambda = 1$, this is no
loss of generality. Then
$$
  e^{h(z)} = \frac{e^z - p(z)}{q(z)}
$$ 
and for $|z| = r$ sufficiently large, some $m \in \Bbb N$, and positive real constants
$C_1, \ldots, C_4$
$$ e^{\operatorname{Re}h(z)} = 
 \lvert e^{h(z)} \rvert \le C_1 (\lvert e^z \rvert + C_2 r^m)
 \le C_1 ( e^r  + C_2 r^m)  \le C_3 e^r 
$$
and therefore
$$
  \operatorname{Re }h(z) \le C_4 +  r
$$
for  $|z| = r > R_0$. So
$$
A(r) :=\max_{|z|=r}\operatorname{Re}h(z) \le  C_4 +  r
$$
Now use 


*

*Coefficients of analytic expansion, or

*Can the real part of an entire function be bounded above by a polynomial?
to estimate the Taylor coefficients of $h$
in terms of $A(r)$, and conclude that $h$ a polynomial
of degree at most one: $h(z) = az + b$. 
Finally, substitute $h$ back into equation $(*)$ and compare the 
growth of the different terms.
