Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that: 
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove the general case and then say that it is true for the specific case, but for the specific case I decided to write out the term and show that they were identical since they were so few.
For the LHS
$$\sum_{k = 0}^{4} (1+x)^k = (1+x)^0 + (1+x)^1 + (1+x)^2 + (1+x)^3 + (1+x)^4$$
$$ = (1) + (1 + x) + (x^2 +2x + 1) + (x^3 + 3x^2 + 3x + 1) + (x^4+4 x^3+6 x^2+4 x+1)$$
$$=5 + 10x + 10x^2 + 5x^3 + x^4$$
For the RHS:
$$\sum_{k=1}^5 \binom{5}{k}x^{k-1} = \binom{5}{1}x^{1-1} + \binom{5}{2}x^{2-1} + \binom{5}{3}x^{3-1} + \binom{5}{4}x^{4-1} + \binom{5}{5}x^{5-1}$$
$$= \frac{5!}{1!4!} x^0 + \frac{5!}{2!3!} x^1 + \frac{5!}{3!2!} x^2 + \frac{5!}{4!1!} x^3 + \frac{5!}{5!0!} x^4$$
$$ = 5 + 10x + 10x^2 + 5x^3 + x^4$$
That completes the first step of the question. So far so good.
For the general case, I started by using the binomial theorem for the binom and then writing out the inner-most sum:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \left( \binom{k}{0}x^0 + \binom{k}{1}x^1 + ... + \binom{k}{k}x^{k}\right)$$
I can imagine that each step in the sum will decide the coefficients for the various powers of x and thus be identical to the RHS of the general case.
However, I run of out steam here and do not at the moment see any obvious way forward. What are some productive approaches for the general case? Am I doing things too complicated?
 A: For $\sum_{k = 0}^{n-1} (1+x)^k$ simplify with the geometric summation formula
For $\sum_{k=1}^n \binom{n}{k}x^{k-1}$ use the binomial theorem
A: Using $a^n-1=(a-1)(\sum_{k=0}^{n-1}a^k)\implies\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}$ for the first equality below (with $a=x+1$), we have
$$
\sum_{k=0}^{n-1}(x+1)^k=\frac{(x+1)^n-1}{x+1-1}=\frac{1}{x}(-1+(x+1)^n)=\frac{1}{x}\left(-1+\sum_{k=0}^n\binom{n}{k}x^k\right)=\frac{1}{x}\sum_{k=1}^n\binom{n}{k}x^k
$$
which simplifies to your RHS.
A: Try induction in $n$, and then use $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $.
A: You can also do a direct proof following your first computations. You can write
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \sum_{i = 0}^{n-1} \binom{k}{i}x^i$$
using the fact that $\binom{k}{i}=0$ when $i$ is greater that $k+1$. Then, if you invert the sums, you find that $$\sum_{k = 0}^{n-1} (1+x)^k =  \sum_{i = 0}^{n-1} \sum_{k = 0}^{n-1}\binom{k}{i}x^i.$$
Then, a classical result (see just after equation (8) here) says that
$$ \sum_{k = 0}^{n-1} \binom{k}{i} = \binom{n}{i+1}.$$
Therefore, you have eventually
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{i = 0}^{n-1}  \binom{n}{i+1} x^i$$ which is exactly what you want.
