Alternative formula for the Bernoulli pmf? If I understand correctly, a Bernoulli pmf just needs to assign a probability $p$ if there is a success $(x=1)$, and $1-p$ otherwise $(x = 0)$. Rather than the usual formula, can't the following function also satisfy this condition and represent a the pmf of a Bernoulli RV?
$
f(x) = xp + (1-x)(1-p).
$
So if $f(x=1) = p, f(x=0) = 1-p.$
What would be wrong in formulating it this way?

There's nothing wrong with it as it evaluates the values, it should evaluate. The usual formulation however uses powers so it becomes a case of binomial distribution with $n=1$ sample size. Recall that the probability mass function of binomial distribution is
$$
{n \choose x} \,p^x (1-p)^{n-x}
$$
Where we have $n$ independent Bernoulli distributed with $x$ successes observed, hence $p^x$, and $n-x$ failures. The ${n \choose x}$ corrects for the fact that we want to account for the successes and failures appearing in any possible combination. With $n=1$ and $x \in \{0,1\}$ it reduces to Bernoulli distribution formulated with the powers
$$
\,p^x (1-p)^{1-x}
$$
