If $n$ & $k$ are positive integers such that $$n \ge \frac{(k)(k+1)}{2}$$ then the number of solutions $(x_1,x_2,...,x_k)$ such that $x_1 \ge 1 , x_2 \ge 2,..., x_k \ge k$ are all integers satisfying $$x_1 + x_2 + ... + x_k = n$$ is?
My attempt at a solution:
The answer equals the coefficient of $x^n$ in the following $$(x+x^2+x^3+....)(x^2+x^3+x^4+...)...(x^k+x^{k+1}...)$$ This equals $$x^{\frac {k(k+1)}{2}}(1+x+x^2+x^3+...)^k$$ This is where I'm stuck. How do I find the coefficient of $x^n$ if I don't know what it equals?
In your question, you got a very nice approach so far. So here is an answer which follows your approach further.
The number of solutions is to be given as a function of k and n. Now with your approach, that number is the number of terms with power $p = n - \frac {k(k+1)}{2}$ in $(1+x+x^2+x^3+...)^k$.
Let $|x| <1$ then the sum in parantheses is (geometric series)
$$ \frac{1}{1-x} $$
So the power you are looking for is given by expansion of
$$ f(x) = \frac{1}{(1-x)^k} $$
into a series about $x=0$, where (Taylor) the coefficient with power $p$ has the factor
$$ \frac{1}{p!} \frac{d^{(p)}f(x)}{dx^{(p)}} |_{x=0} $$
Now taking the derivatives is easy to see:
$$ \frac{d^{(p)}f(x)}{dx^{(p)}} = \frac{k(k+1) \cdots (k+p-1)}{(1-x)^{k+p}} $$
so the factor in the Taylor expansion is
$$ \frac{k(k+1) \cdots (k+p-1)}{p!} = \frac{(k+p-1)!}{p! (k-1)!} = \binom{k+p-1}{k-1} $$
Finally, plugging in $p = n - \frac {k(k+1)}{2}$ from above, we get
$$ \binom{k-1+n - \frac {k(k+1)}{2} }{k-1} $$