Proof of $(a+b)^{n+1}$ I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.
I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.
I however have trouble with the second one, I don't know what to start exactly. Any help/hints?

Note that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$, so

$$\begin{align}
(a+b)^{n+1}&=(a+b)^n(a+b)\\&=\left(\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k}\right)(a+b)\\
&=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\binom{n}{k-1}a^kb^{n-k+1}+\sum_{k=1}^{n}\binom{n}{k}a^kb^{n-k+1}+b^{n+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\left(\binom{n}{k-1}+\binom{n}{k}\right)a^kb^{n-k+1}+b^{n+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^kb^{n-k+1}+b^{n+1}\\&=\sum_{k=0}^{n+1}\binom{n+1}{k}a^kb^{n-k+1}.\end{align}$$

