How may I derive the following identity?
$$ \Big( \sum_{n=0}^\infty c_n x^n \Big)^k = \sum_{n=0}^\infty \tilde{c}_n x^{n-1} $$
$$ \text{where } \ \tilde{c}_n = \sum_{\substack{n_1 \dots n_k \\ n_1+ \cdots+n_k +1 = n}} c_{n_1} \dots c_{n_k} \qquad$$
I have no idea how to proceed. Any suggestion is appreciated.
Reference: Page 4 of this document.
Just sort for powers of $x$: $$ \Big( \sum_{n=0}^\infty c_n x^n \Big)^k = \\ \sum_{n_1=0}^\infty c_{n_1} x^{n_1} \sum_{n_2=0}^\infty c_{n_2} x^{n_2} \cdots \sum_{n_k=0}^\infty c_{n_k} x^{n_k} \\ = \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_k=0}^\infty c_{n_1} c_{n_2} \cdots c_{n_k} x^{n_1 + n_2 + \cdots + {n_k} }\\ = \sum_{n=0}^\infty \tilde{c}_n x^{n}$$
where the $\tilde{c}_n$ are sums over all products $c_{n_1} c_{n_2} \cdots c_{n_k}$ where ${n_1} + {n_2} + \cdots {n_k}$ matches the power $n$.
Now, for your slightly different notation, you can shift $n$ by $1$ in the sum ${n_1} + {n_2} + \cdots {n_k}$ and in the power of $x$. Then you write the power $x^{n-1}$ and correspondingly you must match ${n_1} + {n_2} + \cdots {n_k} = n-1$ or ${n_1} + {n_2} + \cdots {n_k} +1 = n$.