The answer is
There is one valid solution which is $S=4, I=2, N=7, U=8, E=1, V=5, L=9$
Formatting as above
First we note that $N \times N \equiv L (\bmod 10)$ so immediately we cannot have $N=0,1,5,6$ as then $L=N$ which is not permitted. Also important is $I \times N \equiv S (\bmod 10)$ (1)
We can thus consider the problem for different values of $N$. In the following, let $M_i$ represent the $i$th digit in the first multiplication (reading left to right)
$N=2 \Rightarrow L=4 \Rightarrow M_3 + S \equiv I (\bmod 10)$ and we must have $M_3 = S$ but then from (1) above that means that $4S \equiv 2I \equiv S (\bmod 10)$ which means $S=0$ and this cannot be since it is a leading digit.
$N=3 \Rightarrow L=9 \Rightarrow M_3 = S$ and using (1) we must have $6S \equiv 3I \equiv S (\bmod 10)$ which allows $S=2,4,6,8$ paired with $I=4,8,2,6$ respectively. Clearly, since the second multiplication has one less digit than the first, we must have $I < N$ so we can rule out all but the third of these possibilities but then $623 \times 2$ has four digits instead of three so this won't work.
$N=4 \Rightarrow L=6 \Rightarrow M_3 = S+1$ which means that $8S+4 \equiv 4I \equiv S (\bmod 10)$ which means $S=8$ and $I=7$ but this doesn't work since we must have $I<N$.
$N=7 \Rightarrow L=9 \Rightarrow M_3 \equiv S+4 (\bmod 10)$ which means that $14S+28 \equiv 7I \equiv S(\bmod 10)$ or more simply $3S \equiv 2(\bmod 10)$ which gives $S=4$ and $I=2$. This gives a valid solution with additionally $U=8, E=1, V=5$
$N=8 \Rightarrow L=4 \Rightarrow M_3 \equiv S+6 (\bmod 10)$ which means that $16S + 48 \equiv 8I \equiv S (\bmod 10)$ or more simply $5S \equiv 2 (\bmod 10)$ which has no solutions.
Finally, $N=9 \Rightarrow L=1 \rightarrow M_3 \equiv S+8 (\bmod 10)$ which gives us $18S+72 \equiv 9I \equiv S(\bmod 10)$ or more neatly $7S \equiv 8 (\bmod 10)$ which gives $S=4$ and $I=6$ but this gives four digits in the second multiplication instead of three.
Having checked all the cases, we find there is only one valid solution.