I'm currently struggling a bit on this question. I tried induction on the number of parameters.
Induction.
Start $n=0$:
Let $f$ be a boolean function with $0$ parameters.
2 Cases: $f()=0$ or $f()=1$
For case 1 we have a formula like $a \land \lnot a$
Case 2 formula $a \lor \lnot a$
Now the induction step is where i am struggling:
$f(x_1,x_2,...x_n,x_{n+1})=1$ or
$f(x_1,x_2,...x_n,x_{n+1})=0$
Just use the conjunctive normal form, or the disjunctive normal form no induction needed, really. Both obey your requirements, I think.
But if you want induction, try: given $f: \{0,1\}^{n+1} \to \{0,1\}$ define the "sections" $g_0: \{0,1\}^n \to \{0,1\}$ by $g_0(x_1,\ldots,x_n) = f(x_1,\ldots,x_n, 0)$ and $g_1$ similarly with $x_{n+1} = 1$.
By induction represent $g_0(x_1,\ldots, x_n) = \phi_0(x_1, \ldots, x_n)$ using some formula with $\land, \lor$ and $\lnot$ in the terms $x_i$. Ditto with $g_1$ and some $\phi_1$. Then $$f(x_1, \ldots,x_n, x_{n+1}) = (\phi_0(x_0,\ldots,x_n) \land \lnot x_{n+1}) \lor (\phi_1(x_0,\ldots,x_n) \land x_{n+1})$$
which is also of the required form.
