Proving polynomial relationship using combinations Generalize the polynomial relationship $$(1+x)^n=\sum_{k=0}^n x^k {n \choose k}$$ for all positive integers $n$ to $$(a+x)^n$$
My work so far:
I want to use proof by induction to prove this. I am trying to find a polynomial to plug in for $x$ so that this relationship is satisfied. Proving the base case, however, $n=0$ will yield always yield a result of $1=1$. So our base case is satisfied. Additionally, the case for $n=1$ will yield $x+a$, so I must find a polynomial raised to the first power plus one that will result in $x+a$. That polynomial is $x+a-1$. My goal now is to prove this equality. 
$$(a+x)^n=\sum_{k=0}^n (a+x-1)^k {n \choose k}.$$
I've already shown the base case is satisfied; now I will solve for the example for $n+1$ $$(a+x)^{n+1}=\sum_{k=0}^{n+1} (a+x-1)^k {n+1 \choose k}$$ $$(a+x)^n(a+x)=\sum_{k=0}^n (a+x-1)^k {n \choose k} + (a+x-1)^{n+1}{n+1 \choose k}$$.
At this point, I get lost and don't know where to continue from here. I assume there's something I have to do with Pascal's rule, but I don't see where that will lead me. 

You know $$(1+x)^n=\sum_{k=0}^n  {n \choose k}x^k$$
Thus 
$$(a+x)^n = a^n(1+\frac {x}{a})^n=$$
$$  a^n\sum_{k=0}^n{n \choose k} (\frac {x}{a})^k =$$
$$\sum_{k=0}^n{n \choose k} x^k a^{n-k} $$
