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?
 A: I think you need to understand the mathematical induction well. 
You want to show the binomial theorem $(a+b)^n = \sum_{i = 0}^{n} \binom{n}{i}a^ib^{n-i}$, right?  
To do this, You need to assume that this statement is true when $n = N$, then prove for the $n = N+1$ case with your assumption. 
Then, if this is true for $n = 1$, then it is true for $n =2$, and so on for $n = 3,4, \cdots$ . So, for every natural number, the statement is true. 
So, you should assume that $(a+b)^n = \sum\binom{n}{i}a^ib^{n-i}$, and prove $(a+b)^{n+1} = \sum\binom{n+1}{i}a^ib^{n+1-i}$.
I will show you a little example about induction; Prove $2^n > n$ for every natural number. 
First, check the initial case; $2^1 = 2 >1$. 
Now, assume that the statement is true when $n=N$, i.e. $2^N > N$. 
Note that this is not the end of the proof; we did nothing yet. This is just an assumption. With this assumption, we will see what could be happened. 
$2^{N+1} = 2^N \times 2 = 2^N + 2^N > 2N \ge N + 1$. (By our assumption)
This is $n = N+1$ case, so we proved the statement. Now the statement is true for $n = 2$, $n=3$, $\cdots$.
A: Note that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$, so

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

A: Consider,
$$ e^x e^y = e^{x+y}$$
Expand out both sides as power series:
$$ \sum_i \frac{x^i}{i!} \sum_j \frac{y^j}{j!} = \sum_{k=0}^{\infty} \frac{(x+y)^k}{k!} \tag{1}$$
Consider sum on LHS,
$$ \sum_{ i, j \geq 0}^{\infty}\frac{x^i y^j}{i! j!} $$
Now, consider summing the series as sets of homogenous polynomials that is set the condition, $i+j=k$ this makes our sum:
$$ \sum_{k=0}^{\infty} \left(\sum_{i=0}^k \frac{ x^i y^{k-i} }{i! (k-i)!} \right) \tag{2}$$
Equate RHS of (1) to (2):
$$  \sum_{k=0}^{\infty} \left(\sum_{i=0}^k \frac{ x^i y^{k-i} }{i! (k-i)!} \right) = \sum_{k=0}^{\infty} \frac{(x+y)^k}{k!}$$
Rewrite this sum as:
$$ \sum_{k=0}^{\infty} Q_k = \sum_{k=0}^{\infty} P_k$$
Now notice that $Q_k$ is a homogenous polynomial of degree $k$, and $P_k$ is also a homogenous polynomial of degree $k$, hence these two quantities must be equal  Refer
$$ \sum_{i=0}^k \frac{ x^i y^{k-i}}{i!(k-i)!} = \frac{(x+y)^k}{k!}$$
Or rearranging,
$$ \sum_{i=0}^k \binom{k}{i} x^i y^{k-i} = (x+y)^k$$
