Use Rouche`s theorem to prove # of zeros For a fixed $\lambda$ satisfying $\vert\lambda\vert$ < 1, show that $(z - 1)^n e^z + \lambda (z + 1)^n$ has
n zeros in the right half-plane, which are all simple if $\lambda \not=$ 0.
I would really appreciate any help. 

You took a semicircle in the right half plane. Good idea!
$$
(z - 1)^n e^z + \lambda (z + 1)^n=0\iff \left(\frac{z-1}{z+1}\right)^n+\lambda e^{-z}=0,$$
since $z+1\ne 0$ for $z$ with $\operatorname{Re}z>0.$   
Let $f(z)=\left(\frac{z-1}{z+1}\right)^n,$ $g(z)=\lambda e^{-z}.$ 
Take a semicircle in the right half plane with radius $R$ sufficiently large such that $\left(\frac{R-1}{R+1}\right)^n>\lambda$, then
$$
|f(z)|\ge \left(\frac{|z|-1}{|z|+1}\right)^n=\left(\frac{R-1}{R+1}\right)^n>\lambda \quad \text{for } z\,(|z|=R)
$$
and $$
|f(z)|=1\quad \text{for } z=it, t\in \mathbb{R}.$$
So $|f(z)|> \lambda$ on the boundary of the semicircle.
On the contrary 
$$
|\lambda e^{-z}|=\lambda e^{-\operatorname{Re}z}\le \lambda,$$
since $\operatorname{Re}z\ge 0.$
