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I know that the pmf is $$p(x) = p^{k}(1-p)^{1-k}$$ for $k \in \{0,1\}$. Suppose we want to find the joint pdf for i.i.d $X_1, \dots X_n$. Then would if be $$p_{n}(x) = p^{kn} (1-p)^{n(1-k)}$$

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I think you mean $p(k) = p^k (1-p)^{1-k}$, i.e. $p(0) = 1-p$ and $p(1) = p$. Please note that it is a pmf (probability mass function), not a pdf (probability density function). –  Robert Israel May 29 '12 at 18:32
    
$f(k) = p^{nk} (1-p^n)^{1-k}$ for $k \in \{0,1\}$ would be the pmf for the product $X_1 \ldots X_n$. The joint pmf for $X_1,\ldots,X_n$ has to be a function of $n$ variables: $p_n(k_1,\ldots,k_n) = p(k_1) p(k_2) \ldots p(k_n)$ –  Robert Israel May 29 '12 at 18:41
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The product of the $X_i$ is $1$ precisely if all the $X_i$ are $1$. The probability that all the $X_i$ are $1$ is $p^n$. From this you will be able to write down a suitable pdf, since the product of the $X_i$ is itself Bernoulli.

Remarks: $1$. The pdf proposed in the post cannot be correct, since except in a few cases, it cannot be even be a pdf.

$2.$ Your version of the pdf for the individual $X_i$, though perfectly correct, may have led indirectly to the incorrect joint pdf. I think it would have been better to say that $X_i$ is $1$ with probability $p$, and $0$ with probability $1-p$. Your formula for the pdf of $X_i$ is more compact, but perhaps it distances you somewhat from what's really going on.

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