Expansion of $\bigl(\sum x_i\bigr)^4$ Show that $$(\sum_{i=1}^n X_i)^4=\sum_{i=1}^n X_i^4+4\sum_{i\neq j}^n X_i^3X_j+3\sum_{i\neq j}^n X_i^2X_j^2+6\sum_{i\neq j\neq k}^n X_i^2X_jX_k+\sum_{i\neq j\neq k\neq l}^n X_iX_jX_kX_l$$ Please show it step by step.Thanks in advance.

More directly, if you prefer:
$$\begin{align}
\left(\sum_i X_i\right)^4&=\left(\sum_i X_i\right)\left(\sum_j X_j\right)\left(\sum_k X_k\right)\left(\sum_\ell X_\ell\right)\\
&=\sum_i\sum_j\sum_k\sum_\ell X_iX_jX_kX_\ell\\
&=\sum_{i,j,k,\ell} X_iX_jX_kX_\ell\\
\end{align}$$
Now just break apart the 4-tuples according to multiplicities.
$$\begin{align}
&=\sum_{i=j=k=\ell} X_iX_jX_kX_\ell\\
&+\left(\sum_{i=j=k\neq\ell} X_iX_jX_kX_\ell+\sum_{i=j=\ell\neq k} X_iX_jX_kX_\ell+\sum_{i=\ell=k\neq j} X_iX_jX_kX_\ell+\sum_{\ell=j=k\neq i} X_iX_jX_kX_\ell\right)\\
&+\left(\sum_{i=j\neq k=\ell} X_iX_jX_kX_\ell+\sum_{i=k\neq j=\ell} X_iX_jX_kX_\ell+\sum_{i=\ell\neq k=j}X_iX_jX_kX_\ell\right)\\
&+\left(\text{six such things with two equal indices and a third and fourth distinct index}\right)\\
&+\sum_{i, j, k, \ell \text{ distinct}} X_iX_jX_kX_\ell\end{align}$$
