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The other answers so far have given an intuitive explanation. I'd like to show you how the equations work if we model a transformer.
If we simplify the transformer by assuming the no-load resistance drop is very small, then we can say that the induced EMF in the transformer is equal to the applied voltage. If we assume that there is no load on the transformer and we assume that the applied voltage is sinuoidal, the induced EMF is sinusoidal and the flux is sinusoidal, we can say that the induced EMF in the primary is \$e_1=N_1 \frac{d\phi}{dt}\$, where \$e_1\$ is the induced EMF, \$N_1\$ is the number of turns in the primary, and \$\phi\$ is the flux in the core.
As I assumed above, \$\phi\$ is a sinusoid so we can write \$\phi = \phi_{max} sin(\omega t)\$. Then we can say that \$e_1=N_1 \frac{d\phi}{dt} = \omega N_1 \phi_{max} cos(\omega t)\$. If we rearrange that and also remember our assumption that the induced EMF is equal to the applied voltage, we get \$\phi_{max}= \frac{V}{\sqrt2 \pi f N_1}\$.
Basically what this equation says is that our peak flux is proportional to the applied voltage and inversely proportional to the frequency of our applied voltage and the number of turns in the primary of the transformer. The higher your flux is the more steel you need in your transformer in order to keep the flux density at a reasonable level, so that means higher frequency transformers can be smaller.