Proof of $\frac{Y^{\lambda}-\lambda}{\sqrt{\lambda}}\to Z\sim N(0,1)$ in distribution as $\lambda\to\infty$? This is an exercise of the Central Limit Theorem:

Let $Y^{\lambda}$ be a Poisson random variable with parameter $\lambda>0$.
  Prove that $\frac{Y^{\lambda}-\lambda}{\sqrt{\lambda}}\to Z\sim N(0,1)$ in distribution as $\lambda\to\infty$.

I've done that 
$$
Z_n\to Z\sim N(0,1)
$$
in distribution using the CLT, where $Z_n=(Y^n-n)/\sqrt{n}$. Some naive attempt to go is considering 
$$
Y^{n}\leq Y^{\lambda}\leq Y^{n+1}\tag{*}
$$
where $n\leq\lambda\leq n+1$ and somehow use the squeeze theorem. But both (*) and the squeeze theorem in convergence in distribution are NOT justified. How can I go on? Or do I need an alternative direction?
 A: The squeeze theorem in convergence in distribution can be made fully rigorous in the situation you describe--but the shortest proof here might be through characteristic functions. 
Recall that if $Y^\lambda$ is Poisson with parameter $\lambda$, $\varphi_\lambda(t)=\mathbb E(\mathrm e^{\mathrm itY^\lambda})$ is simply $\varphi_\lambda(t)=\mathrm e^{-\lambda(1-\mathrm e^{\mathrm it})}$. Thus $\mathbb E(\mathrm e^{\mathrm itZ^\lambda})=\mathrm e^{-\mathrm it\sqrt{\lambda}}\varphi_\lambda(t/\sqrt{\lambda})=\mathrm e^{-g_\lambda(t)}$ with
$$
g_\lambda(t)=\mathrm it\sqrt{\lambda}+\lambda-\lambda\mathrm e^{\mathrm it/\sqrt{\lambda}}.
$$
Expanding the exponential up to second order yields
$$
g_\lambda(t)=\mathrm it\sqrt{\lambda}+\lambda-\lambda\cdot(1+\mathrm it/\sqrt{\lambda}-t^2/2\lambda)+o(1)\to\tfrac12t^2.
$$
Thus, for every $t$, $\mathbb E(\mathrm e^{\mathrm itZ^\lambda})\to\mathrm e^{-t^2/2}=\mathbb E(\mathrm e^{\mathrm itZ})$ where $Z$ is standard normal, hence $Z^\lambda\to Z$ in distribution.
