1. Define the problem and notation:
Let $p$ be a prime number. We denote by $N(p)$ the number of integers $x$ such that $1 \leq x \leq p$ and $x^x \equiv 1 \pmod{p}$. We aim to prove that there exist constants $c < \frac{1}{2}$ and $p_0 > 0$ such that $N(p) \leq p^c$ for all $p \geq p_0$.

2. Divisors and sets:
For each divisor $d$ of $p-1$, let $A_d$ be the set of integers $1 \leq x \leq p$ such that $\gcd(x, p-1) = d$ and $x^x \equiv 1 \pmod{p}$. This condition implies that $x$ is an $e$-th power modulo $p$, where $e = \frac{p-1}{d}$.

3. Summing over divisors:
We have:
$$N(p) = \sum_{d \mid p-1} |A_d|$$
Note that $|A_d| \leq \min(d, e)$, so $|A_d| \leq \sqrt{p-1}$. Therefore:
$$N(p) \leq \sum_{d \mid p-1} \sqrt{p-1}$$
Since the number of divisors $\tau(n)$ of $n$ satisfies $\tau(n) \ll n^\varepsilon$ for any $\varepsilon > 0$, we get:
$$N(p) \leq p^{1/2} \cdot \tau(p-1) \ll p^{1/2 + \varepsilon}$$
for any $\varepsilon > 0$.

4. Improving the bound:
To improve the bound, we need a good estimate for $|A_d|$ when $d \approx e \approx p^{1/2}$. This is where the sum-product theorem over finite fields is useful.

5. Sum-product theorem application:
The sumset $A_d + A_d$ is contained in the set of multiples of $d$ within $\{1, 2, \ldots, 2p-1\}$, so:
$$|A_d + A_d| \leq 2e$$
Similarly, the product set $A_d \cdot A_d$ is contained in the set of $e$-th powers in $\mathbb{F}_p^\times$, so:
$$|A_d \cdot A_d| \leq d$$
Hence:
$$\max(|A_d + A_d|, |A_d \cdot A_d|) \leq 2 \max(d, e)$$
By the sum-product theorem, the left-hand side is at least $|A_d|^{\frac{5}{4} - \varepsilon}$, implying:
$$|A_d|^{\frac{5}{4} - \varepsilon} \leq 2 \max(d, e)$$
Therefore:
$$|A_d| \ll \max(d, e)^{\frac{4}{5} + \varepsilon}$$

6. **Bounding $|A_d|$:**
For $\min(d, e) \ll p^{4/9}$, we use the original bound to get $|A_d| \ll p^{4/9}$. Otherwise, if $d, e \ll p^{5/9}$, we get:
$$|A_d| \ll p^{4/9 + \varepsilon}$$

7. **Final bound for $N(p)$:**
Summarizing, we have shown that:
$$N(p) \ll p^{4/9 + \varepsilon}$$
Hence, any $c > \frac{4}{9}$ is admissible. This shows that there exist constants $c < \frac{1}{2}$ and $p_0 > 0$ such that $N(p) \leq p^c$ for $p \geq p_0$.

$\blacksquare$