A sequence $a_1,a_2,a_3,\ldots $ of non-negative integers is such that $a_{n+1}$ is the last digit of $a_n^n+a_{n-1}$ for all $n>2$. Is it always true that for some $n_0$ the sequence $a_{n_0},a_{n_0+1},a_{n_0+2},\ldots$ is periodic?
1. **Understanding the Problem:**
   We are given a sequence \(a_1, a_2, a_3, \ldots\) of non-negative integers such that \(a_{n+1}\) is the last digit of \(a_n^n + a_{n-1}\) for all \(n > 2\). We need to determine if the sequence eventually becomes periodic.

2. **Periodic Nature of Last Digits:**
   For any digit \(d\), the sequence of its powers \(d, d^2, d^3, \ldots\) is periodic when considering only the last digit. For example:
   - If \(d = 2\), the sequence of last digits is \(2, 4, 8, 6, 2, 4, \ldots\) with a period of 4.
   - If \(d = 3\), the sequence of last digits is \(3, 9, 7, 1, 3, 9, \ldots\) with a period of 4.
   - If \(d = 7\), the sequence of last digits is \(7, 9, 3, 1, 7, 9, \ldots\) with a period of 4.

   In general, the last digit of \(d^n\) for any digit \(d\) has a periodicity of at most 4.

3. **Least Common Multiple of Periods:**
   Let \(N\) be the least common multiple (LCM) of the periods of the sequences of last digits for all digits \(0, 1, 2, \ldots, 9\). Since the maximum period is 4, \(N = 4\).

4. **Defining the Triple:**
   For every \(n > 2\), define a triple \((r(n, N), a_n, a_{n-1})\), where \(r(n, N)\) is the remainder when \(n\) is divided by \(N\). Since \(N = 4\), \(r(n, 4)\) can be \(0, 1, 2,\) or \(3\).

5. **Finite Number of Triples:**
   There are finitely many possible values for the triple \((r(n, 4), a_n, a_{n-1})\) because:
   - \(r(n, 4)\) can take 4 values: \(0, 1, 2, 3\).
   - \(a_n\) and \(a_{n-1}\) are digits, so each can take 10 values: \(0, 1, 2, \ldots, 9\).

   Therefore, there are \(4 \times 10 \times 10 = 400\) possible triples.

6. **Conclusion:**
   Since there are only 400 possible triples and the sequence is infinite, by the Pigeonhole Principle, some triple must repeat. Once a triple repeats, the sequence from that point onward will be periodic.

\[
\boxed{\text{True}}
\]