Consistent last digit?

The last digit of \(3^3\) is 7.
The last digit of \(3^{3^3} \) is 7.
The last digit of \(3^{3^{3^3}} \) is 7.

The 3 statements above are all true. Is it true that for all integers \(n\geq2\), \[\underbrace{3^{3^{3^{3^{\cdot^{\cdot^3}}}}}}_{n\text{ number of 3's}} \] has a last digit of 7?

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