[proofplan] We prove the result by induction on $n$. The base case $n = 2$ is the key computation: the joint probability $\mathbb{P}(X_1 = x_1, X_2 = x_2)$ factors by independence, allowing the double sum to separate into a product of single sums. [/proofplan]

[step:Prove the base case $n = 2$ by factoring the joint distribution]Let $X_1, X_2$ be independent discrete random variables. By the definition of expectation applied to the product $X_1 X_2$: \begin{align*} \mathbb{E}[X_1 X_2] &= \sum_{x_1} \sum_{x_2} x_1 x_2 \, \mathbb{P}(X_1 = x_1, X_2 = x_2). \end{align*} Since $X_1$ and $X_2$ are independent, $\mathbb{P}(X_1 = x_1, X_2 = x_2) = \mathbb{P}(X_1 = x_1)\,\mathbb{P}(X_2 = x_2)$. Substituting and separating the sums: \begin{align*} \mathbb{E}[X_1 X_2] &= \sum_{x_1} \sum_{x_2} x_1 x_2 \, \mathbb{P}(X_1 = x_1)\,\mathbb{P}(X_2 = x_2) \\ &= \left(\sum_{x_1} x_1 \, \mathbb{P}(X_1 = x_1)\right)\left(\sum_{x_2} x_2 \, \mathbb{P}(X_2 = x_2)\right) \\ &= \mathbb{E}[X_1]\,\mathbb{E}[X_2]. \end{align*} The separation of the double sum into a product of single sums is valid because $x_1 \, \mathbb{P}(X_1 = x_1)$ does not depend on $x_2$ and can be factored out of the inner sum.[/step]

[guided]The double sum $\sum_{x_1}\sum_{x_2} x_1 x_2\,\mathbb{P}(X_1 = x_1)\,\mathbb{P}(X_2 = x_2)$ has the form $\sum_{x_1}\sum_{x_2} f(x_1)\,g(x_2)$ where $f(x_1) = x_1\,\mathbb{P}(X_1 = x_1)$ and $g(x_2) = x_2\,\mathbb{P}(X_2 = x_2)$. Since $f(x_1)$ does not depend on $x_2$, we pull it out of the inner sum: \begin{align*} \sum_{x_1} f(x_1) \sum_{x_2} g(x_2) = \left(\sum_{x_1} f(x_1)\right)\left(\sum_{x_2} g(x_2)\right). \end{align*} This factorisation is the step that requires independence — without it, the joint probability $\mathbb{P}(X_1 = x_1, X_2 = x_2)$ cannot be split into a product, and the double sum does not separate. We also need absolute convergence of the sums $\sum_{x_1} |x_1|\,\mathbb{P}(X_1 = x_1)$ and $\sum_{x_2} |x_2|\,\mathbb{P}(X_2 = x_2)$ (which is guaranteed by the hypothesis that the expectations exist) to justify the rearrangement.[/guided]

[step:Complete the induction for general $n$] Assume the result holds for $n - 1$ independent random variables. Given $X_1, \ldots, X_n$ independent, define $Y = X_1 \cdots X_{n-1}$. By the [Preservation of Independence](/theorems/1116) theorem, $Y = f(X_1, \ldots, X_{n-1})$ and $X_n$ are independent (since $Y$ is a function of $X_1, \ldots, X_{n-1}$, which are independent of $X_n$). Applying the base case to $Y$ and $X_n$: \begin{align*} \mathbb{E}[X_1 \cdots X_n] = \mathbb{E}[Y \cdot X_n] = \mathbb{E}[Y]\,\mathbb{E}[X_n]. \end{align*} By the inductive hypothesis, $\mathbb{E}[Y] = \mathbb{E}[X_1 \cdots X_{n-1}] = \prod_{i=1}^{n-1} \mathbb{E}[X_i]$. Therefore \begin{align*} \mathbb{E}\left[\prod_{i=1}^n X_i\right] = \prod_{i=1}^n \mathbb{E}[X_i]. \end{align*} [/step]