Proving $(x+y)^n = \sum\limits_{k=0}^n \binom{n}{k} x^k y^{n-k}$ I'm reading Serge Lang's 'Analysis I', and there's a problem I cannot figure out how to prove:

Problem: Prove by induction that $$(x+y)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k} . $$

Attempt at proof : I established the base case, which is easily verified. Now I want to prove the inductive step. So assume the assertion holds for $n \geq 1$, and we want to show it also holds for $n + 1$. So we have to proof that: $$(x+y)^{n+1} = \sum_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} x^k y^{n+1-k} . $$
I started with the the LHS and applied the induction hypothesis:
\begin{align*} (x+y)^{n+1} &= (x+y)^n (x+y) \\ & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k} (x+y) \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^{k+1} y^{n-k} + \sum_{k=0}^n  \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k+1} \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \bigg[ x^{k+1} y^{n-k} + x^k y^{n-k+1} \bigg] \end{align*} Now I don't know how to proceed. Any help would be appreciated! 

Hint: $$\begin{align}
\sum_{k=0}^n &\binom n  k \bigg[ x^{k+1} y^{n-k} + x^k y^{n-k+1} \bigg]\\
&=\color{red}{\sum_{k  = 0}^n {n \choose k} x^{k+1}y^{n-k}} + \color{#00A000}{\sum_{k  = 0}^n {n \choose k} x^{k} y^{n-k+1}}\\
&=\color{red}{{n \choose n} x^{n+1} + \sum_{k  = 0}^{n-1} {n \choose k} x^{k+1} y^{n-k}} + \color{#00A000}{{n \choose 0} y^{n+1} + \sum_{k  = 1}^{n} {n \choose k} x^{k} y^{n-k+1}}\\
&=x^{n+1} + y^{n+1} + \color{royalblue}{\sum_{k  = 0}^{n-1} {n \choose k} x^{k+1} y^{n-k}} + \sum_{k  = 1}^{n} {n \choose k} x^{k} y^{n-k+1}\\
&={n+1 \choose 0} x^{n+1} + {n+1 \choose n+1} y^{n+1} + \color{royalblue}{\sum_{k  = 1}^n {n \choose k-1} x^{k} y^{n-k+1}} + \sum_{k  = 1}^n {n \choose k} x^{k} y^{n-k+1}.\\
\end{align}$$
Can you continue?
