(I assume $\mathbb{N}$ to have $0$. That is, $\mathbb{N}:=\mathbb{Z}^+\cup \{0\}$)
Let $n,k$ be a positive integers such that $k\leq n$.
Define $S:=\{\alpha\in \mathbb{N}^n: \sum_{m=1}^n m\alpha_m=n\text{ and } \sum_{m=1}^n \alpha_m = k\}$.
Define $T:=\{\alpha\in (\mathbb{Z}^+)^k: \sum_{m=1}^k \alpha_m=n\}$.
Define $P:=\sum_{\alpha\in S} \frac{n!}{\alpha_1 ! \cdots \alpha_n !} (\frac{X_1}{1!})^{\alpha_1}\cdots (\frac{X_n}{n!})^{\alpha_n}$.
Define $Q:=\frac{1}{k!}\sum_{\alpha\in T} \frac{n!}{\alpha_1 ! \cdots \alpha_k !} X_{\alpha_1}\cdots X_{\alpha_k}$.
How do I prove that $P=Q$?
Some texts use $P$ as the definition of Bell polynomial and some texts use $Q$ as the definition of Bell polynomial. I'm curious whether these two polynomials are identical.
Thank you in advance!
Define $F:=\sum_{m=1}^\infty \frac{X_m}{m!}t^m (\in \mathbb{Q}[X_1,X_2,...][[t]])$.
Define $\Phi:=\sum_{k=0}^\infty \frac{u^kF^k}{k!} (\in \mathbb{Q}[X_1,X_2,...][[t]]][[u]])$.
Let $n$ be a positive integer.
Regard $\Phi$ as a formal power series in $\mathbb{Q}[X_1,X_2,...][[u]][[t]]$.
Then, we have $$\Phi(n)=\sum_{k=0}^n \frac{u^kF^k(n)}{k!}=\sum_{k=1}^n \frac{u^kF^k(n)}{k!}-(*)$$.
Let's expand $(*)$ in two different ways.
First expansion
$$(*)=(\sum_{k=1}^n \frac{u^k F^k}{k!})(n)=(\sum_{k=1}^n \frac{u^k}{k!}\sum_{\alpha_1+\alpha_2+ \cdots=k} \frac{k!}{\alpha_1! \alpha_2!\cdots} (X_1\frac{t}{1!})^{\alpha_1}(X_2\frac{t^2}{2!})^{\alpha_2}\cdots )(n)= (\sum_{k=1}^n \frac{u^k}{k!}\sum_{\alpha_1+\alpha_2+ \cdots=k} \frac{k!}{\alpha_1! \alpha_2!\cdots} [(\frac{X_1}{1!})^{\alpha_1}(\frac{X_2}{2!})^{\alpha_2}\cdots ]t^{\alpha_1+2\alpha_2+\cdots})(n)= \sum_{k=1}^n \frac{u^k}{k!}\sum_{\alpha_1+\cdots +\alpha_n=k, \alpha_1+\cdots + n\alpha_n =n}\frac{k!}{\alpha_1!\cdots \alpha_n!} X_1^{\alpha_1}\cdots X_n^{\alpha_n}$$.
Second expansion
$$(*)= \sum_{k=1}^n \frac{u^k}{k!} \sum_{\alpha_1+\cdots +\alpha_k=n} F(\alpha_1)\cdots F(\alpha_k)=\sum_{k=1}^n \frac{u^k}{k!} \sum_{\alpha_1+\cdots +\alpha_k=n} \frac{X_{\alpha_1}}{\alpha_1!}\cdots \frac{X_{\alpha_k}}{\alpha_k!}$$
Conclusion
$$ \sum_{k=1}^n \frac{u^k}{k!}\sum_{\alpha_1+\cdots +\alpha_n=k, \alpha_1+\cdots + n\alpha_n =n}\frac{k!}{\alpha_1!\cdots \alpha_n!} X_1^{\alpha_1}\cdots X_n^{\alpha_n}=(*)=\sum_{k=1}^n \frac{u^k}{k!} \sum_{\alpha_1+\cdots +\alpha_k=n} \frac{X_{\alpha_1}}{\alpha_1!}\cdots \frac{X_{\alpha_k}}{\alpha_k!}$$.
By comparing the coefficients of $u^k$, we have $P=Q$.