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! 

By induction, illustrating the case $n=3\to n=4$, which easily generalizes:
$$(x+y)\left(\binom33x^3y^0+ \binom32x^2y^1+ \binom31x^1y^2+ \binom30x^0y^3\right)\\=$$
$$\begin{align}\binom33x^4y^0+&\binom32x^3y^1+ \binom31x^2y^2+ \binom30x^1y^3+\\
&\binom33x^3y^1+ \binom32x^2y^2+ \binom31x^1y^3+ \binom30x^0y^4\end{align}\\=$$
$$\binom44x^4y^0+ \binom43x^3y^1+ \binom42x^2y^2+ \binom41x^1y^3+ \binom40x^0y^4\\$$
$$1\ 3\ 3\ 1-\\-1\ 3\ 3\ 1\\1\ 4\ 6\ 4\ 1$$
The base case is just
$$\binom00x^0y^0$$
$$1$$
