Here's a slightly fleshed-out version of Nicholas Kuhn's argument. I wasn't able to verify the exact statement he uses, but a variant thereof.
For a spectrum $X$, let $f(X)$ be the maximal $k \in \mathbb N$ such that $Sq^k$ acts nontrivially on $H^\ast(X)$ (note that we're not using $Sq^{2^k}$). If $X$ is finite, then $f(X) < \infty$. Then clearly
$f(X \oplus Y) = \max(f(X),f(Y))$
$f(\Sigma X) = f(X)$
Let us show that:
- $f(X^n) = n f(X)$$f(X_1\wedge X_2) = f(X_1) + f(X_2)$
By the Cartan formula, $Sq^k(x_1 \otimes \cdots \otimes x_n) = \sum_{k_1 + \dots + k_n = k} S^{k_1}(x_1) \otimes \cdots Sq^{k_n}(x_n)$$Sq^k(x_1 \otimes x_2) = \sum_{k_1 + k_2 = k} S^{k_1}(x_1) \otimes Sq^{k_2}(x_2)$. If $k > nf(X)$$k > f(X_1) + f(X_2)$, then in each summand, at least one tensor factor has $Sq^{k_i}$ acting for $k_i > f(X)$$k_i > f(X_i)$, so $Sq^k$ vanishes. Thus $f(X^n) \leq n f(X)$$f(X_1 \wedge X_2) \leq f(X_1)+f(X_2)$. If $k = n f(X)$$k = f(X_1) + f(X_2)$, then every summand has a tensor factor for which $Sq^{k_i}$ acts with $k_i > f(X)$$k_i > f(X_i)$ except for $k_1 = k_2 = \cdots = f(X)$$k_i = f(X_i)$, so $Sq^k = Sq^{f(X)} \otimes \cdots \otimes Sq^{f(X)}$$Sq^k = Sq^{f(X_1)} \otimes Sq^{f(X_2)}$. So if $x \in H^\ast(X)$ is$x_i \in H^\ast(X_i)$ are such that $Sq^{f(X)}(x) \neq 0$$Sq^{f(X_i)}(x_i) \neq 0$, then $Sq^k(x \otimes \cdots \otimes x) \neq 0$$Sq^k(x_1 \otimes x_2) \neq 0$. Thus $f(X^n) \geq n f(X)$$f(X_1 \wedge X_2) \geq f(X_1) + f(X_2)$.
Now suppose that $X$ is integral, i.e. $X^n \cong \oplus_{i=0}^{n-1}\oplus_j a_{ij} \Sigma^j X^i$. Then by the above twothree formulas, we have $n f(X) = f(X^n) = f(\oplus_{i=0}^{n-1} \oplus_j a_{ij} \Sigma^j X^i) = \max_{i=0}^{n-1} i f(X) = (n-1) f(X)$. Therefore $f(X) = 0$, i.e. the Steenrod algebra acts trivially on $H^\ast(X)$.