Past exam of the logic and set theory course of the University of Oxford / 2010 / B / 5 / d

Yes, with the stated strict decrease and positive coefficients. In any displayed Cantor normal form, the first exponent is the unique $\delta$ with ${\omega }^{\delta }\le \alpha <{\omega }^{\delta +1}$. The leading coefficient is the unique positive integer $n$ withThe remaining tail is the unique $\zeta <{\omega }^{\delta }$ satisfying $\alpha ={\omega }^{\delta }n+\zeta$, by strict monotonicity of ordinal addition in the right operand.

Thus any two such representations have the same first exponent and coefficient and the same remainder. Repeat the argument on that remainder. Their finite lists agree term by term and reach zero at the same step, so the lengths agree as well. Allowing zero coefficients or repeated exponents would spoil this normalized uniqueness, which is why those conditions are required.