Suppose that $Y_{\lambda}=^{d}P(\lambda)$. Prove that $[Y_{\lambda}-\lambda]/{\sqrt{\lambda}}\to^{d}N(0,1)$

Question

Suppose that $Y_{\lambda}=^{d}P(\lambda)$. Prove that $[Y_{\lambda}-\lambda]/{\sqrt{\lambda}}\to^{d}N(0,1)$ when $\lambda \to \infty$ using characteristic functions. So $$\phi(t)=\sum_{k=0}^{\infty} \frac{e^{it\lambda}}{\sqrt{\lambda}}\left(e^{-\lambda}\frac{\lambda^k}{k!}-\lambda\right)$$ where $\phi(t)$ stands for characteristic function. So $$\phi(t)=\frac{1}{\sqrt{\lambda}}\sum_{k=0}^{\infty} {e^{it\lambda}}\left(e^{-\lambda}\frac{\lambda^k}{k!}-\lambda\right)=\frac{1}{\sqrt{\lambda}}[\sum_{k=0}^{\infty} {e^{it\lambda}}e^{-\lambda}\frac{\lambda^k}{k!}-\sum_{k=0}^{\infty}e^{it\lambda}\lambda]$$ Then $$\frac{1}{\sqrt{\lambda}}[\sum_{k=0}^{\infty} {e^{it\lambda}}e^{-\lambda}\frac{\lambda^k}{k!}-\sum_{k=0}^{\infty}e^{it\lambda}\lambda]=\frac{1}{\sqrt{\lambda}}\left[ {e^{\lambda(e^{it-1})}}-\lambda\sum_{k=0}^{\infty}e^{it\lambda}\right].$$ Stuck at this point.