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R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

Note:
5 bytes are wasted because in rank function the default handling strategy in case of ties is "average their ranks" instead of any of the other possibilities... argh!