Range of entire function. Picard's theorem $f(z)$$=$ 2$z^{14}$$cos^2z$ 

$f(z)$ is entire function. I want to figure out if $f$($\Bbb C$)$=$$\Bbb C$ or there's a point $a$ such that $f$($\Bbb C$)$=$$\Bbb C$\ {$a$}. So essentially I want to use Little Picard Theorem here.

I thought about using the reflection principle. Let $f(z)$$=$$w$.  So here $f$($\bar{z}$)=$\bar{f(z)}$. This means that if $w$ if this exceptional point then $\bar{w}$ is too. I guess this implies that exceptional points can only be located on the real line?

Can someone please correct me if I am wrong. Any help with solving this problem would be appreciated.
Along the real axis, we have $f(z) = f(x) = 2x^{14}\cos(x)^2 \ge 0$.
Notice $f(0) = 0$ and $\lim\limits_{n\to\infty} f(n\pi) = 2(n\pi)^{14} \to \infty$. Apply IVT along positive real axis, we have


$$f([0,\infty)) \supset [0,\infty)\quad\implies\quad f(\mathbb{R}) = [0,\infty)$$

Along the imaginary axis, we have $f(z) = f(iy) = -2y^{14}\cosh(y)^2 \le 0$.
Notice $f(i0) = 0$ and $\lim\limits_{y\to+\infty} f(iy) = -\infty$. Apply IVT along positive imaginary axis, we have
$$f(i[0,\infty)) \supset (-\infty,0]\quad\implies\quad f(i\mathbb{R}) = (-\infty,0]$$


Combine them, we have $f(\mathbb{R} \cup i\mathbb{R}) = \mathbb{R}$.
As you have already known, by reflection principle, if $f(z)$ avoids any number $\alpha$, $\alpha$ need to be real. Since this have been ruled out by above argument, $f(z)$ does not avoid any number and $f(\mathbb{C}) = \mathbb{C}$.
Notice that $f(z)$ is the product of two entire functions $g(z)=2z^{14}$ and $h(z)=cos^2z$ where $g$ is onto by fundamental theorem of algebra, so $f(\mathbb C)=\mathbb C$.